ฝึกอินทิเกรตกัน

[กระบี่เดียวดายแสวงพ่าย]

น่าสนุกดีครับ ขอโจทย์อีกข้อด้วยนะครับ

[nooonuii]

เติมโจทย์ให้ครับ

จงหาค่าของ $$\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx$$

[-InnoXenT-]

ข้อความเดิมเขียนโดยคุณ nooonuii

เติมโจทย์ให้ครับ

จงหาค่าของ $$\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx$$

ให้ $u = \tan{\frac{x}{2}}$ ดังนั้น

$dx = \frac{2du}{1+u^2}$

$$\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx$$

$$=\int \Big(\dfrac{\frac{2u}{1+u^2}+2(\frac{1-u^2}{1+u^2})}{3(\frac{2u}{1+u^2})+4(\frac{1-u^2}{1+u^2})}\Big) \frac{2du}{1+u^2}$$

$$=2\int \dfrac{u^2-u-1}{(1+u^2)(2u+1)(u-2)} \, du$$

จัด Partial fraction แล้ว integrate จะได้

$$\frac{2}{25}(-\ln{(u^2+1)}+\ln{(u-2)}+\ln{(2u+1)}+11\tan^{-1}{u}) + c$$

แล้วจัดรูปธรรมดา ก็จะได้

$$\frac{11x}{25}+\frac{2}{25}\ln{\left| 9\sin{x}+12\cos{x} \right|} + c $$

ฝากโจทย์หน่อยครับ

$$\int \sqrt{\cot{x}} \,dx$$

[Ne[S]zA]

http://www.wolframalpha.com/input/?i...%2B4cosx%29+dx

[nooonuii]

ข้อความเดิมเขียนโดยคุณ nooonuii

เติมโจทย์ให้ครับ

จงหาค่าของ $$\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx$$

ผมมีอีกสองวิธีที่ต่างกัน แต่ขอเอาอันนี้มาลงไว้ละกัน

$\sin{x}+2\cos{x}=\dfrac{11}{25}(3\sin{x}+4\cos{x})+\dfrac{2}{25}(3\cos{x}-4\sin{x})$

ดังนั้น

$\displaystyle{\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx=\dfrac{11}{25}\int 1\, dx + \dfrac{2}{25}\int\dfrac{3\cos{x}-4\sin{x}}{3\sin{x}+4\cos{x}}\, dx}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{11x}{25}+\dfrac{2}{25}\ln{|3\sin{x}+4\cos{x}|}+C$

[-InnoXenT-]

ข้อความเดิมเขียนโดยคุณ nooonuii

ผมมีอีกสองวิธีที่ต่างกัน แต่ขอเอาอันนี้มาลงไว้ละกัน

$\sin{x}+2\cos{x}=\dfrac{11}{25}(3\sin{x}+4\cos{x})+\dfrac{2}{25}(3\cos{x}-4\sin{x})$

ดังนั้น

$\displaystyle{\int\Big(\dfrac{\sin{x}+2\cos{x}}{3\sin{x}+4\cos{x}}\Big)\, dx=\dfrac{11}{25}\int 1\, dx + \dfrac{2}{25}\int\dfrac{3\cos{x}-4\sin{x}}{3\sin{x}+4\cos{x}}\, dx}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{11x}{25}+\dfrac{2}{25}\ln{|3\sin{x}+4\cos{x}|}+C$

ผมว่า คุณลืม คูณ 3 นะ ^^"

ฝากข้อของผมด้วยนะ XD

[กระบี่เดียวดายแสวงพ่าย]

ของคุณnooonuii เกือบถูกครับ
แต่ของคุณInnoXenT ตรวจสอบโดยการดิฟแล้วไม่น่าถูกครับ

[Little Penguin]

มันไม่เกือบถูกหรอกครับ แต่มันถูกเลยแหละครับ

ตอบคุณ -InnoXenT- นะครับ

3 ที่คูณอยู่ เราดึงออกมาจากใน log ได้ แล้วมันจะเป็นค่าคงที่ครับ ดังนั้นคำตอบถูกทั้งคู่ครับ

[กระบี่เดียวดายแสวงพ่าย]

เอ แต่ผมว่าลองดิฟแล้วไม่กลับมาเป็นฟังก์ชันเหมือนโจทย์ตอนแรกเลยครับ

[gnopy]

กระทู้นี้เล่นกันข้ามปีเลยนะครับ

มาแปะข้อง่ายๆให้ครับ

1. $\int \frac{1}{e^x+1}dx$

2. $\int x^{x^{x^{x^{^{.}}}}}dx$

[-InnoXenT-]

ข้อความเดิมเขียนโดยคุณ Little Penguin

มันไม่เกือบถูกหรอกครับ แต่มันถูกเลยแหละครับ

ตอบคุณ -InnoXenT- นะครับ

3 ที่คูณอยู่ เราดึงออกมาจากใน log ได้ แล้วมันจะเป็นค่าคงที่ครับ ดังนั้นคำตอบถูกทั้งคู่ครับ

จะว่าไป มันก็ใช่แฮะ ;D ขอโทษครับ

ข้อ 1. ของคุณ gnopy

จะได้ $\ln{\frac{e^x}{e^x+1}}$

ข้อ 2. ทำไงเนี่ย ไม่รู้ - -a

[Little Penguin]

ข้อคุณ -InnoXenT- เราให้ $u=\sqrt{\cot{x}}$ จะได้อินทิกรัลเป็น

$\displaystyle\int\sqrt{\cot{x}}\,dx=-\int\frac{2u^2}{u^4+1}\,du$

สังเกตว่า $\displaystyle \frac{2u^2}{u^4+1}=\frac{\sqrt{2}}{2} \left(\frac{u}{u^2-\sqrt{2}u+1} -\frac{u}{u^2+\sqrt{2}u+1}\right)$

ซึ่ง $\displaystyle \frac{u}{u^2-\sqrt{2}u+1} =\frac{1}{2}\cdot\frac{2u-\sqrt{2}}{u^2-\sqrt{2}u+1} +\frac{\sqrt{2}}{\left(\sqrt{2}u-1\right)^2+1}$

และ $\displaystyle \frac{u}{u^2+\sqrt{2}u+1} =\frac{1}{2}\cdot\frac{2u+\sqrt{2}}{u^2+\sqrt{2}u+1} -\frac{\sqrt{2}}{\left(\sqrt{2}u+1\right)^2+1}$

ดังนั้น $\displaystyle -\int\frac{2u^2}{u^4+1}\,du =-\frac{\sqrt{2}}{2}\left(\int\frac{1}{2}\cdot\frac{2u-\sqrt{2}}{u^2-\sqrt{2}u+1}\,du +\int\frac{\sqrt{2}}{\left(\sqrt{2}u-1\right)^2+1}\,du -\int\frac{1}{2}\cdot\frac{2u+\sqrt{2}}{u^2+\sqrt{2}u+1}\,du +\int\frac{\sqrt{2}}{\left(\sqrt{2}u+1\right)^2+1}\,du\right)$

$\displaystyle =-\frac{\sqrt{2}}{2}\left(\frac{1}{2}\ln\left|u^2-\sqrt{2}u+1\right| +\tan^{-1}\left(\sqrt{2}u-1\right) -\frac{1}{2}\ln\left|u^2 +\sqrt{2}u+1\right|+\tan^{-1}\left(\sqrt{2}u+1\right)\right)+C$

$\therefore\displaystyle\int\sqrt{\cot{x}}\,dx =-\frac{1}{2\sqrt{2}}\left( \ln\left|\cot{x}-\sqrt{2\cot{x}}+1\right| -\ln\left|\cot{x} +\sqrt{2\cot{x}}+1\right|+2\tan^{-1}\left(\sqrt{2\cot{x}}-1\right) +2\tan^{-1}\left(\sqrt{2\cot{x}}+1\right)\right)+C$ $\square$

[-InnoXenT-]

แล้วก็ฝากข้อใหม่อีกหลายๆข้อด้วยครับ

1. $$\int \dfrac{e^x+1}{e^{2x}+1} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...Fpostcount=270

2. $$\int \sqrt{x+\sqrt{x}} \, dx$$

เฉลย

let $u = \sqrt{x}$ ---> $2udu = dx$

$$\int \sqrt{x+\sqrt{x}} \, dx = \int 2u\sqrt{u^2+u}\, du$$

$$= \int 2u\sqrt{(u+\frac{1}{2})^2-\frac{1}{4}} \, du$$

let $u+\frac{1}{2} = \frac{1}{2}\sec{v}$

$$= \frac{1}{2}\int (\sec{v}-1)\frac{1}{2}\tan^2{v}\sec{v} \, dv$$

$$= \frac{1}{4}\int \sec^2{v}\tan^2{v}\, dv - \frac{1}{4}\int \sec{v}\tan^2{v} \, dv$$

$$= \frac{1}{4}\int \tan^2{v} \, d(\tan{v}) - \frac{1}{4}\int \sec^3{v} \, dv + \frac{1}{4}\int \sec{v} \, dv$$

$$=\frac{1}{12}\tan^3{v}-\frac{1}{8}\tan{v}\sec{v} + \frac{1}{8}\ln{(\sec{v}+\tan{v})} + C$$

then, substitute it back

$$=\sqrt{x+\sqrt{x}}(\frac{2}{3}x-\frac{1}{6}\sqrt{x}-\frac{1}{4})+\frac{1}{8}\ln{(2\sqrt{x+\sqrt{x}}+2\sqrt{x}+1)} + C$$

3. $$\int \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...Fpostcount=270

4. $$\int \sin{(\ln{x})} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...Fpostcount=270

5. $$\int \frac{\sqrt{1+\ln{x}}}{x\ln{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...Fpostcount=270

6. $$\int x^4(\ln{x})^2 \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...Fpostcount=270

7. $$\int \frac{\ln{x}}{(1+x^2)^{3/2}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=290

8. $$\int \frac{x^2\sin^{-1}{x}}{\sqrt{1-x^2}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=282

9. $$\int \frac{x^2+1}{x\sqrt{x^4+1}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=281

10. $$\int \frac{\sin^2{x}}{1+\sin^2{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=280

11. $$\int e^{\sin{x}}\frac{x\cos^3{x}-\sin{x}}{\cos^2{x}} \, dx$$

เฉลย

$$\int e^{\sin{x}}\frac{x\cos^3{x}-\sin{x}}{\cos^2{x}} \, dx = \int xe^{\sin{x}}\cos{x} \, dx - \int \frac{e^{\sin{x}}\sin{x}}{\cos^2{x}} \, dx$$

For the first integral use integration by part by using $u = x$ and $dv = e^{\sin{x}}\cos{x}dx$, we get

$$\int xe^{\sin{x}}\cos{x} \, dx = xe^{\sin{x}} - \int e^{\sin{x}} \, dx + c$$

For the second integral use integration by part by using $u = e^{\sin{x}}$ and $dv = \sec{x}\tan{x} dx$ , we get

$$\int \frac{e^{\sin{x}}\sin{x}}{\cos^2{x}} \, dx = e^{\sin{x}}\sec{x} - \int e^{\sin{x}} \, dx + c$$

hence,

$$\int xe^{\sin{x}}\cos{x} \, dx - \int \frac{e^{\sin{x}}\sin{x}}{\cos^2{x}} \, dx = xe^{\sin{x}} - \int e^{\sin{x}} \, dx - (e^{\sin{x}}\sec{x} - \int e^{\sin{x}} \, dx) + c$$

$$= e^{\sin{x}}(x-\sec{x}) + c$$

12. $$\int \frac{\sqrt[3]{x}}{x(\sqrt{x}+\sqrt[3]{x})} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=275

13. $$\int e^{-x}\sin{(x+\frac{\pi}{4})} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=289 ขออนุญาตเอาของ R.Wasutharat นะครับ อ่านง่ายกว่า

14. $$\int_{0}^{\frac{\pi}{4}} \frac{x\cos{(5x)}}{\cos{x}} \, dx$$

เฉลย

just use the fact that $\cos{(5x)} = 16\cos^5{x}-20\cos^3{x}+5\cos{x}$ and continue the integration by part. The answer is $\frac{1}{32}(8-8\pi-\pi^2)$

15. $$\int_{0}^{\pi} \frac{x\sin{x}}{1+\cos^2{x}} \, dx$$

เฉลย

let $x = \pi - u$ and $I = \int_{0}^{\pi} \frac{x\sin{x}}{1+\cos^2{x}} \, dx$

we got $$I = \pi \int_{0}^{\pi} \frac{\sin{x}}{1+\cos^2{x}} \, dx - I$$

$$I = \frac{\pi}{2} \int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2{x}}\, dx$$

$$= -\frac{\pi}{2}\int_{0}^{\pi}\frac{d(\cos{x})}{1+\cos^2{x}}$$

$$= -\frac{\pi}{2}\left[ \tan^{-1}{(\cos{x})}\right]_{0}^{\pi}$$

$$I = -\frac{\pi}{2}(-\frac{\pi}{4}-\frac{\pi}{4})) = \frac{\pi^2}{4}$$

16. $$\int_{e^{e^e}}^{e^{e^{e^e}}} \frac{1}{(x\ln{x})(\ln{(\ln{x})})(\ln{(\ln{(\ln{x})})})} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=274

17. $$\int \frac{\ln{x}}{(x+1)^2} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=279

18. $$\int_{0}^{\frac{\pi}{2}} \frac{\sin^3{x}}{\sin{x}+\cos{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=285

19. $$\int_{0}^{\frac{\pi}{2}} \frac{\sin{(2553x)}}{\sin{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=300
http://www.mathcenter.net/forum/show...&postcount=302 อีกหนึ่งวิธี

20. $$\int \frac{dx}{\sin{x}\sqrt{1-\cos{x}}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=292

21. $$\int \tan{\theta}\sqrt{\cos{\theta}}\ln{(\cos{\theta})} \, d\theta$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=291

22. $$\int_{0}^{a} \sqrt{2ax-x^2} \, dx (a > 0) $$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=283

23. $$\sum_{k = 1}^{2010} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{\sqrt{1+k}-x} \, dx$$

เฉลย

Consider the integrand $$\int_{-1}^{1} \frac{\sqrt{1-x^2}}{\sqrt{1+k}-x} \, dx$$

$$= \int_{-1}^{1} \frac{\sqrt{1-x^2}}{\sqrt{1+k}-x}\frac{\sqrt{1+k}+x}{\sqrt{1+k}+x} \, dx$$

$$= \int_{-1}^{1}\frac{\sqrt{1-x^2}\sqrt{1+k} + x\sqrt{1-x^2}}{1+k-x^2}\, dx$$

$$= \int_{-1}^{1}\frac{\sqrt{1-x^2}\sqrt{1+k}}{1+k-x^2} \, dx + \int_{-1}^{1}\frac{x\sqrt{1-x^2}}{1+k-x^2} \, dx$$

For the second integral let $v = -x$ ,then we know that it equals to $0$

$$\int_{-1}^{1} \frac{\sqrt{1-x^2}}{\sqrt{1+k}-x} \, dx = \int_{-1}^{1}\frac{\sqrt{1-x^2}\sqrt{1+k}}{1+k-x^2} \, dx$$

Let $x = \sin{u}$ ---> $dx = \cos{u}du$

$$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos{u}\sqrt{1-\sin^2{u}}\sqrt{1+k}}{k+\cos^2{u}} \, du$$

$$= \sqrt{k+1}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos{(2u)}}{2k+1+\cos{(2u)}} \, du$$

let $t = \tan{u}$

$$= \sqrt{k+1}\int_{-\infty}^{\infty} \frac{1+\frac{1-t^2}{1+t^2}}{2k+1+\frac{1-t^2}{1+t^2}} \, \frac{dt}{1+t^2}$$

$$= \sqrt{k+1}\int_{-\infty}^{\infty} (\frac{1}{1+t^2} - \frac{2k}{2k+2+2kt^2}) \, dt$$

$$= \sqrt{k+1}(\pi - \sqrt{\frac{k}{k+1}}\pi)$$

Hence,

$$\sum_{k = 1}^{2010} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{\sqrt{1+k}-x} \, dx = \sum_{k=1}^{2010} = \pi \sum_{k=1}^{2010} (\sqrt{k+1}-\sqrt{k}) = (\sqrt{2011} - 1)\pi$$

24. $$\int_{0}^{\pi} \cos{(2x)}\cos{(2^2x)}\cos{(2^3x)} ... \cos{(2^{2010}x)} \, dx$$

เฉลย

$$= \int_{-\pi}^{\pi} (\frac{\sin{2^2x}}{2\sin{2x}})(\frac{\sin{2^3x}}{2\sin{2^2x}})(\frac{\sin{2^4x}}{2\sin{2^3x}}) ... (\frac{\sin{2^{2011}x}}{2\sin{2^{2010}x}}) \, dx$$

$$= \frac{1}{2^{2010}}\int_{-\pi}^{\pi} \frac{\sin{2^{2011}x}}{\sin{2x}} \, dx$$

$$=\frac{1}{2^{2010}} \int_{-\pi}^{\pi} (\frac{\sin{(2(2^{2010}-1)+2)x}}{\sin{2x}}) \, dx$$

$$=\frac{1}{2^{2010}} \int_{-\pi}^{\pi}(\frac{\sin{2x}\cos{2(2^{2010}-1)x}+\cos{2x}\sin{2(2^{2010}-1)x}}{\sin{2x}} \, dx$$

For the first integral it's always $= 0$

$$=\frac{1}{2^{2010}} \int_{-\pi}^{\pi} \frac{\cos{2x}\sin{2(2^{2010}-1)x}}{\sin{2x}} \, dx$$

$$= \int_{-\pi}^{\pi} \frac{\cos{2x}(\sin{2x}\cos{2(2^{2010}-2)x}+\cos{2x}\sin{2(2^{2010}-2)x})}{2^{2010}\sin{2x}} \, dx$$

For the first term it's always $= 0$

hence,

$$= \int_{-\pi}^{\pi} \frac{\cos^{2^{2010}-1}{x}}{2^{2010}} \, dx = 0$$

25. $$\int \frac{x^2}{(x^2-4)\sin{x}+4x\cos{x}} \, dx$$

เฉลย

$$\int \frac{x^2}{(x^2-4)\sin{x}+4x\cos{x}} \, dx = \int \frac{x^2}{x^2\sin{x}-4\sin{x}+4x\cos{x}} \, dx$$

$$= \int \frac{x^2}{2x^2\sin{\frac{x}{2}}\cos{\frac{x}{2}}-4x\sin^2{\frac{x}{2}}+4x\cos^2{\frac{x}{2}}-8\cos{\frac{x}{2}}\sin{\frac{x}{2}}} \, dx$$

$$= \frac{1}{2} \int \frac{x^2}{x^2\sin{\frac{x}{2}}\cos{\frac{x}{2}}-2x\sin^2{\frac{x}{2}}+2x\cos^2{\frac{x}{2}}-4\cos{\frac{x}{2}}\sin{\frac{x}{2}}} \, dx$$

$$= \frac{1}{2} \int \frac{x^2}{(x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}})(x\cos{\frac{x}{2}}-2\sin{\frac{x}{2}})}$$

$$= \frac{1}{2} \int \frac{x^2\cos{\frac{x}{2}}+x^2\sin{\frac{x}{2}}-2x\sin{\frac{x}{2}}\cos{\frac{x}{2}}+2x\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{(x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}})(x\cos{\frac{x }{2}}-2\sin{\frac{x}{2}})}$$

$$= \frac{1}{2} \int \frac{x\cos{\frac{x}{2}}}{x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}}} + \frac{x\sin{\frac{x}{2}}}{x\cos{\frac{x}{2}}-2\sin{\frac{x}{2}}} \, dx$$

$$= \int \frac{d(x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}})}{x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}}} + \int \frac{d(x\cos{\frac{x}{2}}-2\sin{\frac{x}{2}})}{x\cos{\frac{x}{2}}-2\sin{\frac{x}{2}}}$$

$$= \ln{(\frac{x\sin{\frac{x}{2}}+2\cos{\frac{x}{2}}}{x\cos{\frac{x}{2}}-2\sin{\frac{x}{2}}})} + C$$

26. $$\int \left| a\cos{(nx)}+b\sin{(nx)} \,\right| \, dx (n = 0,1,2,3,...)$$

เฉลย

if $n = 0$ the integrand will be equal to $\left| a \,\right| x + c $
$$= \int \left| a\cos{(nx)}+b\sin{(nx)} \,\right| \, dx = \sqrt{a^2+b^2} \left| \frac{a}{\sqrt{a^2+b^2}}\cos{(nx)}+\frac{b}{\sqrt{a^2+b^2}}\sin{(nx)} \,\right| \, dx$$

Let $\cos{u} = \frac{b}{\sqrt{a^2+b^2}}$ and $n \not= 0$

$$= \frac{\sqrt{a^2+b^2}}{n} \int \left| \sin{(nx+u)} \,\right| \, d(nx+u)$$

$$= -\frac{\sqrt{a^2+b^2}}{n}\left| \cos{nx+u}\,\right| + C = -\frac{\sqrt{a^2+b^2}}{n}\left| \cos{(nx+\arccos{(\frac{b}{\sqrt{a^2+b^2}})})}\,\right| + C$$

27. $$\int_{0}^{\frac{\pi}{2}} \frac{\sin^2{x}\cos{x}}{\sin{x}+\cos{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=300

28. $$\int x^2\cos^{-1}{x} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=277

29. $$\int_{0}^{\pi} \ln{(1+\cos{x})} \, dx$$

เฉลย

$$\int_{0}^{\pi} \ln{(1+\cos{x})} \, dx = \int_{0}^{\pi} \ln{(1-\cos{x})} \, dx = I$$

$$2I = 2\int_{0}^{\pi} \ln{(\sin{x})} \, dx = 4\int_{0}^{\frac{\pi}{2}} \ln{(\sin{x})}\, dx$$

$$I = 2\int_{0}^{\frac{\pi}{2}} \ln{(\sin{x})}\, dx = 2\int_{0}^{\frac{\pi}{2}} \ln{(\cos{x})}\, dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \ln{(\sin{x})} \, dx + \int_{0}^{\frac{\pi}{2}} \ln{(\cos{x})} \, dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \ln{(\sin{2x})} - \ln{2} \, dx$$

$$I = \frac{1}{2}I - \frac{\pi}{2}\ln{2}$$

$$I = -\pi \ln{2}$$

30. $$\int x^5\tan^{-1}{x^2} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=293

31. $$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \, dx$$

เฉลย

Let $x = \frac{\pi}{2} -u$ So the intergrand will be

$$\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos{u}}}{\sqrt{\sin{u}}+\sqrt{\cos{u}}} \, du = I = \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \, dx$$

$$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \, dx = \frac{\pi}{2}$$

$$I = \frac{\pi}{4}$$

32. $$\int_{-1}^{1} \frac{\sin{x}}{1+x^6} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=276

33. $$\int_{0}^{\pi} \ln{(1+b\cos{x})} \, dx$$ (b is constant.)
เฉลย : http://www.mathcenter.net/forum/show...&postcount=301

34. $$\int_{0}^{\infty} \frac{e^{-5x} - e^{-6x}}{x} \, dx $$

เฉลย

$$\int_{0}^{\infty} \frac{e^{-5x} - e^{-6x}}{x} \, dx = \int_{0}^{\infty} \int_{-6x}^{-5x}\frac{e^y}{x} \, dydx $$

$$= \int_{0}^{-\infty}\int_{-\frac{y}{5}}^{-\frac{y}{6}} \frac{e^y}{x} \, dxdy$$

$$= \int_{0}^{-\infty} e^y\ln{\frac{5}{6}} \, dy$$

$$= \ln{\frac{6}{5}}$$

35. $$\int_{e}^{e^2} \frac{1+(\ln{x})(\ln{(\ln{x})})}{\ln{x}}\, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=298

36. $$\int \frac{x}{\sqrt{x^4+4x^3-6x^2+4x+1}} \, dx$$

เฉลย

$$= \int \frac{dx}{\sqrt{(x+\frac{1}{x})^2 + 4(x+\frac{1}{x})-8}} = I$$

Let $u = \frac{1}{x}$

$$I = \int -\frac{du}{u^2\sqrt{(u+\frac{1}{u})^2 + 4(u+\frac{1}{u})-8}}$$

$$2I = \int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x})^2 + 4(x+\frac{1}{x})-8}$$

Let $t = x+\frac{1}{x}$ --> $dt = (1-\frac{1}{x^2}) dx$

$$I = \frac{1}{2}\int\frac{dt}{\sqrt{t^2+4t-8}}$$

$$= \frac{1}{2}\int\frac{dt}{\sqrt{(t+2)^2-12}}$$

$$= \frac{1}{2}\ln{\left| x+\frac{1}{x}+2+\sqrt{(x+\frac{1}{x}+2)^2-12} \,\right| }$$

37. $$\int \frac{1}{x^2+2ax+b} \, dx (a^2>b) $$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=286

38. $$\frac{1}{\ln{\pi}}\int_{\frac{-\pi}{11}}^{\frac{\pi}{11}} \tan{(x^{41}+\frac{x^5}{\pi}-x^3+x)} \, dx$$

เฉลย

Since it is odd function, so the integrand will equal to zero.

39. $$\int_{2}^{4} \frac{\sqrt{\ln{(9-x)}}}{\sqrt{\ln{(9-x)}}+\sqrt{\ln{(3+x)}}}\, dx$$

เฉลย

Let $$I = \int_{2}^{4} \frac{\sqrt{\ln{(9-x)}}}{\sqrt{\ln{(9-x)}}+\sqrt{\ln{(3+x)}}}\, dx$$

Let $3+t = 9-x$

$$I = \int_{2}^{4} \frac{\sqrt{\ln{(3+t)}}}{\sqrt{\ln{(9-t)}}+\sqrt{\ln{(3+t)}}}\, dt = \int_{2}^{4} \frac{\sqrt{\ln{(9-x)}}}{\sqrt{\ln{(9-x)}}+\sqrt{\ln{(3+x)}}}\, dx$$

$$2I = \int_{2}^{4} \, dx$$

$$I = 1$$

40. $$\int (\sqrt{\tan{x}}-\sqrt{\cot{x}}) \, dx$$

เฉลย

Let $u^2 = \tan{x}$

$$\int (\sqrt{\tan{x}}-\sqrt{\cot{x}}) \, dx = \int \frac{2u^2-2}{u^4+1} \, du$$

$$= 2\int\frac{1-\frac{1}{u^2}}{u^2+\frac{1}{u^2}} \, du$$

Let $v = u + \frac{1}{u}$

$$= 2\int \frac{dv}{v^2-2} = \frac{1}{\sqrt{2}}\ln{\left|\frac{\sqrt{2}-v}{\sqrt{2}+v}\,\right|}+c$$

$$= \frac{1}{\sqrt{2}}\ln{\left|\frac{\sqrt{2}-\left|\tan{x} \,\right| - \left| \cot{x}\,\right| }{\sqrt{2}+\left|\tan{x} \,\right| + \left| \cot{x}\,\right| }\,\right|} + c$$

41. Let $x^y = y^x$, evaluate $$\int \frac{\ln{y}-\frac{y}{x}}{\ln{x}-\frac{x}{y}} \, dx$$

เฉลย

$$x^y = y^x$$

$$y\ln{x} = x\ln{y}$$

$$\frac{d}{dx} (y\ln{x}) = \frac{d}{dx}(x\ln{y})$$

$$\frac{dy}{dx}\ln{x}+\frac{y}{x} = \frac{dy}{dx}(\frac{x}{y}) + \ln{y}$$

$$\frac{dy}{dx}=\frac{\ln{y}-\frac{y}{x}}{\ln{x}-\frac{x}{y}}$$

$$\int \frac{\ln{y}-\frac{y}{x}}{\ln{x}-\frac{x}{y}} \, dx = \int \frac{dy}{dx} \, dx$$

$$= \int \, dy = y + C$$

42. $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}+3(\cos{(2x)})(\sqrt{\sin{x}}-\sqrt{\cos{x}})}{\sqrt{\sin{(2x)}}}\, dx$$

เฉลย

Consider $\frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}+3(\cos{(2x)})(\sqrt{\sin{x}}-\sqrt{\cos{x}})}{\sqrt{\sin{(2x)}}}$

$$= \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}+\cos{2x}\sqrt{\sin{x}}-\cos{2x}\sqrt{\cos{x}}}{\sqrt{\sin{(2x)}}}+\frac{2(\sqrt{\sin{x}}-\sqrt{\cos{x}})\cos{(2x)}}{\sqrt{\sin{(2x)}}}$$

$$= \frac{(1+\cos{2x})\sqrt{\sin{x}}+(1-\cos{2x})\sqrt{\cos{x}}}{2\sin{x}\cos{x}}\sqrt{\sin{2x}}+\frac{2(\sqrt{\sin{x}}-\sqrt{\cos{x}})\cos{(2x)}}{\sqrt{\sin{(2x)}}}$$

$$= \frac{2\cos^2{x}\sqrt{\sin{x}}+2\sin^2{x}\sqrt{\cos{x}}}{2\sin{x}\cos{x}}+\frac{2(\sqrt{\sin{x}}-\sqrt{\cos{x}})\cos{(2x)}}{\sqrt{\sin{(2x)}}}$$

$$= \sqrt{\sin{2x}}(\frac{\cos{x}}{\sqrt{\sin{x}}}+\frac{\sin{x}}{\sqrt{\cos{x}}})+\frac{2(\sqrt{\sin{x}}-\sqrt{\cos{x}})\cos{(2x)}}{\sqrt{\sin{(2x)}}}$$

$$= 2\sqrt{\sin{2x}}\frac{d}{dx}(\sqrt{\sin{x}}-\sqrt{\cos{x}}) + (\sqrt{\sin{x}}-\sqrt{\cos{x}})\frac{d}{dx}(2\sqrt{\sin{2x}})$$

$$= \frac{d}{dx}(2\sqrt{\sin{2x}}(\sqrt{\sin{x}}-\sqrt{\cos{x}}))$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}+3(\cos{(2x)})(\sqrt{\sin{x}}-\sqrt{\cos{x}})}{\sqrt{\sin{(2x)}}}\, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{d}{dx}(2\sqrt{\sin{2x}}(\sqrt{\sin{x}}-\sqrt{\cos{x}})) \, dx$$

$$= \left[ 2\sqrt{\sin{2x}}(\sqrt{\sin{x}}-\sqrt{\cos{x}}) \,\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \sqrt{3}-\sqrt{\sqrt{3}}$$

43. $$\int \arctan{\frac{1}{x}} dx$$

เฉลย

Let $u = \arctan{\frac{1}{x}}$ and $dv = dx$ the integral will be

$$= x\arctan{\frac{1}{x}} + \int \frac{x}{1+x^2} \, dx = x\arctan{\frac{1}{x}} +\frac{1}{2}\ln{(1+x^2)}+C$$

44. $$\int_{0}^{\frac{\pi}{2}} (x\cot{x})^2 \, dx$$

เฉลย

$$= \int_{0}^{\frac{\pi}{2}} x^2(\csc^2{x}-1) \, dx$$

$$-\int_{0}^{\frac{\pi}{2}} x^2 d(\cot{x}) - \int_{0}^{\frac{\pi}{2}} x^2 \, dx$$

For the first integral, let $dv = d(\cot{x})$ and $u = x^2$, second integral equal to $\frac{\pi^{3}}{24}$

$$= -\left[ \left[x^2\cot{x}\,\right]_{0}^{\frac{\pi}{2}} -2\int_{0}^{\frac{\pi}{2}} x\cot{x} \, dx \,\right] - \frac{\pi^3}{24}$$

$$= 2\int_{0}^{\frac{\pi}{2}} x\cot{x} \, dx - \frac{\pi^3}{24}$$

$$= 2\int_{0}^{\frac{\pi}{2}} x \, d(\ln{\sin{x}}) - \frac{\pi^3}{24}$$

Let $u =x$ and $dv = d(\ln{\sin{x}})$

$$= 2\left[ \left[ x\ln{\sin{x}}\,\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \ln{\sin{x}} \, dx \,\right] -\frac{\pi^3}{24}$$

$$= 2(-\int_{0}^{\frac{\pi}{2}} \ln{(\sin{x})}\, dx) -\frac{\pi^3}{24}$$

$$= \pi \ln{2} - \frac{\pi^3}{24}$$

PS. you can know the proof of $$\int_{0}^{\frac{\pi}{2}} \ln{\sin{x}}\, dx$$ from question number 29.

45. $$\int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}} \sin{x}\cos^{-1}{\frac{\cos{a}}{\sin{x}}} \, dx$$ ($a$ is an constant.)

เฉลย

$$I = \int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}} \sin{x}\cos^{-1}{\frac{\cos{a}}{\sin{x}}} \, dx$$

$$= -\int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}} \cos^{-1}{(\frac{\cos{a}}{\sin{x}})} d(\cos{x})$$

by part $u = \cos^{-1}{(\frac{\cos{a}}{\sin{x}})}$ and $dv = d(\cos{x})$ ,we get

$$= -\left[ \cos{x}\cos^{-1}{(\frac{\cos{a}}{\sin{x}})} \,\right]_{\frac{\pi}{2}-a}^{\frac{\pi}{2}}+ \int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}} \frac{\cos^2{x}\sin{x}dx}{(1-\cos^2{x})\sqrt{\sin^2{a}-\cos^2{a}}} $$

$$= \int_{\frac{\pi}{2}-a}^{\frac{\pi}{2}} \frac{\cos^2{x}\sin{x}dx}{(1-\cos^2{x})\sqrt{\sin^2{a}-\cos^2{a}}}$$

Let $\cos{x} = \sin{a}\sin{t}$

$$=\cos{a}\int_{0}^{\frac{\pi}{2}} \frac{(\sin{a}\sin{t})^2dt}{1-(\sin{a}\sin{t})^2}$$

$$=-\cos{a}\int_{0}^{\frac{\pi}{2}}dt + \cos{a}\int_{0}^{\frac{\pi}{2}}\frac{dt}{(1-\sin{a}\sin{t})(1+\sin{a}\sin{t})}$$

$$= -\frac{\pi}{2}\cos{a}+\frac{\cos{a}}{2}\int_{0}^{\frac{\pi}{2}}\frac{dt}{1+\sin{a}\sin{t}}+\frac{\cos{a}}{2}\int_{0}^{\frac{\pi}{ 2}}\frac{dt}{1-\sin{a}\sin{t}}$$

Both integral, $u = \tan{\frac{t}{2}}$

First Integral

$$\int_{0}^{\frac{\pi}{2}}\frac{dt}{1+\sin{a}\sin{t}} = \int_{0}^{\frac{\pi}{2}} \frac{2du}{(u^2+1)(\frac{2u\sin{a}}{u^2+1}+1)}$$

$$=\int_{0}^{\frac{\pi}{2}} \frac{2du}{2u\sin{a}+u^2+1}$$

$$=2\int_{0}^{\frac{\pi}{2}}\frac{du}{(\sin{a}+u)^2+(1-\sin^2{a})}$$

and the rest is easy.

Second Integral

$$\int_{0}^{\frac{\pi}{2}}\frac{dt}{1-\sin{a}\sin{t}} = \int_{0}^{\frac{\pi}{2}} \frac{2du}{(u^2+1)(-\frac{2u\sin{a}}{u^2+1}+1)}$$

$$=\int_{0}^{\frac{\pi}{2}} \frac{2du}{-2u\sin{a}+u^2+1}$$

$$=2\int_{0}^{\frac{\pi}{2}}\frac{du}{(\sin{a}-u)^2+(1-\sin^2{a})}$$

and the rest is easy.

$$=-\frac{\pi}{2}\cos{a}+(\tan^{-1}{(\frac{1+\sin{a}}{\cos{a}})}-a)+(\tan^{-1}{(\frac{1-\sin{a}}{\cos{a}})}+a) $$

$$= \frac{\pi}{2}(1-\cos{a})$$

46. $$\int_{0}^{\frac{\pi}{4}} \frac{\ln{(\cot{x})}(\sin{(2x)})^{n-1}}{(\sin^n{x}+\cos^n{x})^2} \, dx$$ (n is an integer greater than 1.)

เฉลย

$$= -\int_{0}^{\frac{\pi}{4}}\frac{\ln{(\tan{x})}(2\sin{x}\cos{x})^{n-1}}{(\sin^n{x}+\cos^n{x})^2} \, dx$$

$$= -\frac{2^{n-1}}{n^2}\int_{0}^{\frac{\pi}{4}} \frac{\ln{(\tan^n{x})}n\tan^{n-1}{x}\sec^2{x}}{(1+\tan^n{x})^2} \, dx$$

Let $t = \tan^n{x}$ ---> $dt = n\tan^{n-1}{x}\sec^2{x}dx$

$$= -\frac{2^{n-1}}{n^2}\int_{0}^{1}\frac{\ln{t}}{(t+1)^2} \, dt$$

By part by $u = \ln{t}$ and $dv = \frac{dt}{(1+t)^2}$

$$= -\frac{2^{n-1}}{n^2}(\left[ \frac{\ln{t}}{t+1} \,\right]_{0}^{1} - \int_{0}^{1} \frac{1}{t(1+t)}\, dt )$$

$$= \frac{2^{n-1}}{n^2}(\lim_{t \to \infty} \frac{t\ln{t}}{t+1}+\ln{2}) = \frac{2^{n-1}}{n^2}\ln{2}$$

47. $$\int_{0}^{\pi} \frac{x\sin{(2x)}\sin{(\frac{\pi}{2}\cos{x})}}{2x-\pi} \, dx$$

เฉลย

$$I = \int_{0}^{\pi} \frac{x\sin{(2x)}\sin{(\frac{\pi}{2}\cos{x})}}{2x-\pi} \, dx$$

Let $x =\pi - u$

$$I = \int_{0}^{\pi}\frac{(\pi-u)\sin{(2\pi-2u)}\sin{(\frac{\pi}{2}\cos{\pi -u})}}{2(\pi-u)-\pi} \, du$$

$$= \int_{0}^{\pi} \frac{(u-\pi)\sin{(2u)}\sin{(\frac{\pi}{2}\cos{u})}}{\pi-2u} \, du$$

$$2I = \int_{0}^{\pi} \sin{(2x)}\sin{(\frac{\pi}{2}\cos{x})} \, dx$$

$$I = \int_{0}^{\pi}\sin{x}\cos{x}\sin{(\frac{\pi}{2}\cos{x})}\, dx$$

Let $v = \frac{\pi}{2}\cos{x}$

$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{4}{\pi^2}v\sin{v} \, dv$$

$$= \frac{4}{\pi^2}\left[ \sin{x}-x\cos{x}\,\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{8}{\pi^2}$$

48. $$\int_{e^{\sqrt{2}}}^{e^2} (\ln{(\ln{x})}+\frac{1}{\ln{x}}) \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=297

49. $$\int_{0}^{1} \ln{(\sqrt{1+x}+\sqrt{1-x})} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=303

50. $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(\pi - 4\theta)\tan{\theta}}{1-\tan{\theta}} \, d\theta$$

เฉลย

Let $\theta = x+\frac{\pi}{4}$

$$= \int_{-\frac{\pi}{2}}^{0} \frac{(\pi -4x-\pi)(\frac{\tan{x}+1}{1-\tan{x}})}{1-(\frac{1+\tan{x}}{1-\tan{x}})} \, dx$$

$$=\int_{-\frac{\pi}{2}}^{0} \frac{4x(1+\tan{x})}{2\tan{x}}$$

$$= \int_{-\frac{\pi}{2}}^{0} 2x\cot{x}\, dx+\int_{-\frac{\pi}{2}}^{0} 2x \, dx$$

$$= \frac{\pi^2}{4}-\frac{\pi}{2}\ln{2}$$

For the methor $\int_{-\frac{\pi}{2}}^{0} x\cot{x} \, dx$ we can see from question 44.

51. $$\int_{0}^{1} \frac{1-x^2}{1+x^2} \frac{dx}{\sqrt{1+x^4}}$$

เฉลย

$$= -\int_{0}^{1}\frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}\frac{dx}{\sqrt{x^2+\frac{1}{x^2}}}$$

Let $u=\frac{1}{x}+x$

$$= \int_{2}^{\infty} \frac{du}{u\sqrt{u^2-2}}$$

Let $u = 2\sec{v}$

$$= \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2}} \, dv$$

$$= \frac{\pi}{2\sqrt{2}}$$

52. $$\int \sin{x}\sqrt{\sin{(2x)}} \, dx$$

เฉลย

Let $x = \frac{\pi}{4}-u$

$$= \frac{1}{\sqrt{2}}\int (-\cos{u}+\sin{u})\sqrt{\cos{2u}}\, du$$

$$= \frac{1}{\sqrt{2}}(\int \sin{u}\sqrt{2\cos^2{u}-1} \, du - \int \cos{u}\sqrt{1-2\sin^2{u}})$$

$$= \frac{1}{2}( - \int \sqrt{2\cos^2{u}-1} \, d(\sqrt{2}\cos{u}) - \int \sqrt{1-2\sin^2{u}}\, d(\sqrt{2}\sin{u}))$$

$$= \frac{1}{2}\cos{x}\sqrt{\sin{2x}}+\frac{1}{4}\ln{\left| \cos{x}+\sin{x}+\sqrt{\sin{2x}}\,\right| } -\frac{1}{4}\sin^{-1}{(\cos{x}-\sin{x})}$$

53. $$\int_{0}^{\infty} (\frac{\tan^{-1}{x}}{x})^3 \, dx$$

เฉลย

Let $x = \tan{u}$

$$= \int_{0}^{\frac{\pi}{2}} (\frac{u}{\tan{u}})^3\sec^2{u} \, du$$

$$= -\frac{1}{2} \int_{0}^{\frac{\pi}{2}} u^3 \, d(\cot^2{u})$$

$$= -\frac{1}{2}(\left[u^3\cot^2{u} \,\right]_{0}^{\frac{\pi}{2}} -\int_{0}^{\frac{\pi}{2}} 3u^2\cot^2{u} \, du)$$

$$= 0 +\frac{3}{2}\int_{0}^{\frac{\pi}{2}}(u\cot^2{u})\, du$$

$$= \frac{3\pi}{2}\ln{2}-\frac{\pi^3}{16}$$

For the second integral , we can see from question 45.

54. $$\int_{0}^{\frac{\pi}{2}} \cos^{n}{x}\cos{(nx)} \, dx$$

เฉลย

$$I_n = \int_{0}^{\frac{\pi}{2}} \cos^{n}{x}\cos{(nx)} \, dx$$

$$= \frac{1}{n}\int_{0}^{\frac{\pi}{2}} \cos^{n}{x} \, d(\sin{(nx)})$$

$$= \frac{1}{n}(\left[ \cos^n{x}\sin{(nx)} \,\right]_{0}^{\frac{\pi}{2}} + n\int_{0}^{\frac{\pi}{2}}\sin{(nx)}\cos^{n-1}{x}\sin{x}\, dx)$$

$$I_n = \int_{0}^{\frac{\pi}{2}} \cos^{n-1}{x}\sin{x}\sin{nx} \, dx$$

$$2I_n = \int_{0}^{\frac{\pi}{2}} \cos^{n-1}{x}(\cos{x}\cos{nx}+\sin{x}\sin{nx}) \, dx$$

$$2I_n = I_{n-1}$$

$$I_1 = \int_{0}^{\frac{\pi}{2}}\cos^2{x} \, dx = \frac{\pi}{4}$$

$$I_n = \frac{\pi}{2^{n+1}}$$

55. $$\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{\sqrt{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=315

56. $$\int_{0}^{\infty} \frac{e^{-t}-2e^{-3t}+e^{-5t}}{t^2} dt$$
เฉลย : http://www.mathcenter.net/forum/show...3&postcount=52

57. $$\int_{0}^{\pi} \frac{dx}{e^{2x}+e^{2\pi}}$$
เฉลย : http://www.mathcenter.net/forum/show...2&postcount=31 In Question 76

58. $$\int_{0}^{\infty} \tan^{-1}{(\frac{k}{x})^2} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...4&postcount=43

59. $$\int \frac{dx}{\sin^3{x}+\cos^3{x}}$$

เฉลย

Consider $\sin^3{x}+\cos^3{x} = (\sin{x}+\cos{x})(1-\sin{x}\cos{x})$

$$= \sqrt{2}(\sin{(x+\frac{\pi}{4})}-\frac{1}{2}\sin{(x+\frac{\pi}{4})}\sin{2x})$$

$$= \sqrt{2}(\sin{(x+\frac{\pi}{4})}-\frac{1}{4}(\cos{(x-\frac{\pi}{4})}-\cos{(3x+\frac{\pi}{4})}))$$

$$=\sqrt{2}(\sin{(x+\frac{\pi}{4})}-\frac{1}{4}\cos{(\frac{\pi}{2}+(x-\frac{3\pi}{4}))}+\cos{(\frac{\pi}{2}+3x-\frac{\pi}{4})})$$

$$=\frac{1}{2\sqrt{2}}(4\sin{(x+\frac{\pi}{4})}+\sin{(x-\frac{3\pi}{4})}-\sin{(3x-\frac{\pi}{4})})$$

$$=\frac{1}{2\sqrt{2}}(4\sin{x}\cos{\frac{\pi}{4}}+4\cos{x}\sin{\frac{\pi}{4}}+\sin{x}\cos{\frac{3\pi}{4}}-\sin{\frac{\pi}{4}}\cos{x}-\sin{3x}\cos{\frac{\pi}{4}}+\cos{3x}\sin{\frac{\pi}{4}})$$

$$=\frac{1}{2\sqrt{2}}(4\sin{x}\cos{\frac{\pi}{4}}+4\cos{x}\sin{\frac{\pi}{4}}-\sin{x}\cos{\frac{\pi}{4}}-\sin{\frac{\pi}{4}}\cos{x}+\sin{3x}\cos{\frac{3\pi}{4}}+\cos{3x}\sin{\frac{\pi}{4}})$$

$$=\frac{1}{2\sqrt{2}}(3\sin{(x+\frac{\pi}{4})}+\sin{(3x+\frac{3\pi}{4})})$$

Let $u = x+\frac{\pi}{4}$

$$=\frac{1}{2\sqrt{2}}(3\sin{u}+\sin{3u})$$

$$=\frac{1}{2\sqrt{2}}(2\sin{2u}\cos{u}+2\sin{u})$$

$$=\frac{1}{\sqrt{2}}\sin{u}(2\cos^2{u}+1)$$

$$=\frac{1}{\sqrt{2}}\sin{u}(\sqrt{3}-\sqrt{2}\sin{u})(\sqrt{3}+\sqrt{2}\sin{u})$$

$$\int \frac{dx}{\sin^3{x}+\cos^3{x}} = \sqrt{2}\int \frac{du}{\sqrt{2}\sin{u}(\sqrt{3}-\sqrt{2}\sin{u})(\sqrt{3}+\sqrt{2}\sin{u})}$$

Let $t = \tan{\frac{u}{2}}$

$$= \frac{1}{\sqrt{2}}\int \frac{(t+1)^2}{3t^5-2t^3+3t}$$

$$= \frac{1}{\sqrt{2}}(\int \frac{8t}{3(3t^4-2t^2+3)} + \frac{1}{3t} \, dt)$$

$$= \frac{8}{3\sqrt{2}}\int \frac{t}{(3t^4-2t^2+3)} \, dt + \frac{1}{3\sqrt{2}} \ln{t}$$

For the first integral, Let $v = t^2$ we get

$$=\frac{4}{3\sqrt{2}}\int \frac{dv}{3v^2-2v+3}$$

$$=\frac{4}{3\sqrt{2}}\int \frac{dv}{((\sqrt{3}v-\frac{1}{\sqrt{3}})^2+\frac{8}{3})}$$

$$=\frac{1}{3}\tan^{-1}{(\frac{1}{2\sqrt{2}}(3v-1))}$$

$$\int \frac{dx}{\sin^3{x}+\cos^3{x}} = \frac{1}{3}\tan^{-1}{(\frac{1}{2\sqrt{2}}(3v-1))}+\frac{1}{3\sqrt{2}}\ln{\left| t\,\right| }+C$$

$$= \frac{1}{3}\tan^{-1}{(\frac{1}{2\sqrt{2}}(3(\tan{\frac{x+\frac{\pi}{4}}{2}})^2-1))}+\frac{1}{3\sqrt{2}}\ln{\left| (\tan{\frac{x+\frac{\pi}{4}}{2}})^2\,\right| }+C$$

60. $$\int_{0}^{\infty} \frac{\sin{x}}{x}(\frac{1-e^{-x}}{x} - e^{-x}) \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...3&postcount=79

61. $$\int_{0}^{\infty} \int_{0}^{\infty} (\frac{e^{-x}-e^{-y}}{x-y})^2\, dx dy$$
เฉลย : http://www.mathcenter.net/forum/show...1&postcount=89

62. $$\int_{0}^{\infty} \frac{\tan^{-1}{(\pi x)}-\tan^{-1}{x}}{x} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...2&postcount=95

63. $$\int_{0}^{\infty} \frac{dx}{(1+x^r)(1+x^2)}$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=306

64. $$\int \frac{\cos^2{x}}{e-\cos^2{x}} \, dx$$
เฉลย : http://www.mathcenter.net/forum/show...&postcount=310

65. $$\int_{0}^{\frac{\pi}{2}} \cos{(\pi\sin^2{x})} \, dx$$

เฉลย

$$I = \int_{0}^{\frac{\pi}{2}} \cos{(\pi\sin^2{x})} \, dx$$

Let $u = \frac{\pi}{2}-x$

$$I = \int_{0}^{\frac{\pi}{2}} \cos{(\pi\cos^2{x})} \, dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \cos{(\pi-\pi\sin^2{x})} \, dx$$

$$I = \int_{0}^{\frac{\pi}{2}} -\cos{(\pi\sin^2{x})} \, dx$$

$$I = -I$$

$$I = 0$$

66. $$\int_{0}^{\frac{\pi}{2}} \sec{x}\ln{(1+\sin{a}\cos{x})} \, dx$$

เฉลย

$$I(a) = \int_{0}^{\frac{\pi}{2}} \sec{x}\ln{(1+\sin{a}\cos{x})} \, dx$$

$$I'(a) = \cos{a}\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin{a}\cos{x}}$$

$$=\cos{a}\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin{a}(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}})}$$

$$= 2\cos{a}\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{2}\sec^2{\frac{x}{2}}dx}{(1-\sin{a})\tan^2{\frac{x}{2}}+1+\sin{a}}$$

$$=\frac{\cos{a}}{1-\sin{a}}\int_{0}^{\frac{\pi}{2}}\frac{\frac{1}{2}\sec^2{\frac{x}{2}}dx}{\tan^2{\frac{x}{2}}+(\frac{1+\sin{a}}{\cos{a}})^2}$$

Let $t = \tan{\frac{x}{2}}$

$$I'(a) = \frac{\cos{a}}{1-\sin{a}}\int_{0}^{1}\frac{dt}{t^2+(\frac{1+\sin{a}}{\cos{a}})^2}$$

$$= \frac{2\cos^2{a}}{1-\sin^2{a}}\left[ \tan^{-1}{\frac{t\cos{a}}{1+\sin{a}}} \,\right]_{0}^{1}$$

$$I'(a) = 2\tan^{-1}{(\frac{\cos{a}}{1+\sin{a}})}$$

$$I'(a) = 2\tan^{-1}{(\frac{1-\tan^2{\frac{a}{2}}}{(1+\tan{\frac{a}{2}})^2})}$$

$$= 2\tan^{-1}{(\frac{1-\tan{\frac{a}{2}}}{1+\tan{\frac{a}{2}}})}$$

$$=2\tan^{-1}{(\tan{(\frac{\pi}{4}-\frac{a}{2})})}$$

$$I'(a) = 2\tan^{-1}{(\tan{(\frac{\pi}{4}-\frac{a}{2})})}$$

$$I(a) = \frac{\pi}{2}a-\frac{\pi^2}{4}+C$$

$$I(0) = \int_{0}^{\frac{\pi}{2}} \sec{x}\ln{(1)} \, dx = 0$$

$$I(a) = \frac{\pi}{2}a-\frac{\pi^2}{4}$$

67. $$\int \frac{\ln{(\cos{x}+\sqrt{\cos{(2x)}})}}{\sin^2{x}} \, dx$$

เฉลย

Let $u=\ln{(\cos{x}\sqrt{\cos{2x}})}$ and $dv = \csc^2{x} \, dx$

$$= -\cot{x}\ln{(\cos{x}+\sqrt{\cos{2x}})} - \int \frac{-\sin{x}-\frac{\sin{2x}}{\sqrt{\cos{2x}}}}{\cos{x}+\sqrt{\cos{2x}}}\cot{x}\, dx$$

$$A = \int \frac{-\sin{x}-\frac{\sin{2x}}{\sqrt{\cos{2x}}}}{\cos{x}+\sqrt{\cos{2x}}}\cot{x}\, dx$$

$$A = -\int \frac{\sin{x}\sqrt{\cos{2x}}+\sin{2x}}{\sqrt{\cos{2x}}(\cos{x}+\sqrt{\cos{2x}})}\, dx$$

$$A = -\int \frac{\cos{x}(\sqrt{\cos{2x}}+2\cos{x})}{\sqrt{\cos{2x}}(\cos{x}+\sqrt{\cos{2x}})} \, dx$$

$$A = -\int \frac{\cos{x} dx}{\cos{x}+\sqrt{\cos{2x}}}-2\int \frac{\cos^2{x}dx}{\sqrt{\cos{2x}}(\cos{x}+\sqrt{\cos{2x}})}$$

$$A = -\int \frac{\cos{x} dx}{\cos{x}+\sqrt{\cos{2x}}} -2\int\frac{\cos{x}dx}{\sqrt{\cos{2x}}}+2\int\frac{\cos{x}dx}{\cos{x}+\sqrt{\cos{2x}}}$$

$$= \int\frac{\cos{x}dx}{\cos{x}+\sqrt{\cos{2x}}}-2\int\frac{\cos{x}}{\sqrt{\cos{2x}}}dx$$

First Integral

$$\int\frac{\cos{x}dx}{\cos{x}+\sqrt{\cos{2x}}} =\int\frac{\cos{x}dx}{\cos^2{x}-\cos{2x}}(\cos{x}-\sqrt{\cos{2x}})$$

$$=\int \frac{\cos{x}(\cos{x}-\sqrt{\cos{2x}})}{\sin^2{x}} \, dx$$

$$=\int\cot^2{x}\, dx - \int\frac{\cos{x}}{\sin^2{x}}\sqrt{\cos{2x}} \, dx$$

$$=-\cot{x}+\csc{x}\sqrt{\cos{2x}}+2\int\frac{\cos{x}}{\sqrt{\cos{2x}}}\,dx$$

$$A =-\cot{x}+\csc{x}\sqrt{\cos{2x}}$$

$$\int \frac{\ln{(\cos{x}+\sqrt{\cos{(2x)}})}}{\sin^2{x}} \, dx = -\cot{x}\ln{(\cos{x}+\sqrt{\cos{2x}})}+\cot{x}-\csc{x}\sqrt{\cos{2x}} $$

บางข้อ ก็เอามาจาก ในเว็บนี้แหละครับ โจทย์ผมหมดสต็อกละ ไว้จะหามาใหม่ ^^"

[-SIL-]

ข้อความเดิมเขียนโดยคุณ Little Penguin

ข้อคุณ -InnoXenT- เราให้ $u=\sqrt{\cot{x}}$ จะได้อินทิกรัลเป็น

$\displaystyle\int\sqrt{\cot{x}}\,dx=-\int\frac{2u^2}{u^4+1}\,du$

สังเกตว่า $\displaystyle \frac{2u^2}{u^4+1}=\frac{\sqrt{2}}{2} \left(\frac{u}{u^2-\sqrt{2}u+1} -\frac{u}{u^2+\sqrt{2}u+1}\right)$

ซึ่ง $\displaystyle \frac{u}{u^2-\sqrt{2}u+1} =\frac{1}{2}\cdot\frac{2u-\sqrt{2}}{u^2-\sqrt{2}u+1} +\frac{\sqrt{2}}{\left(\sqrt{2}u-1\right)^2+1}$

และ $\displaystyle \frac{u}{u^2+\sqrt{2}u+1} =\frac{1}{2}\cdot\frac{2u+\sqrt{2}}{u^2+\sqrt{2}u+1} -\frac{\sqrt{2}}{\left(\sqrt{2}u+1\right)^2+1}$

ดังนั้น $\displaystyle -\int\frac{2u^2}{u^4+1}\,du =-\frac{\sqrt{2}}{2}\left(\int\frac{1}{2}\cdot\frac{2u-\sqrt{2}}{u^2-\sqrt{2}u+1}\,du +\int\frac{\sqrt{2}}{\left(\sqrt{2}u-1\right)^2+1}\,du -\int\frac{1}{2}\cdot\frac{2u+\sqrt{2}}{u^2+\sqrt{2}u+1}\,du +\int\frac{\sqrt{2}}{\left(\sqrt{2}u+1\right)^2+1}\,du\right)$

$\displaystyle =-\frac{\sqrt{2}}{2}\left(\frac{1}{2}\ln\left|u^2-\sqrt{2}u+1\right| +\tan^{-1}\left(\sqrt{2}u-1\right) -\frac{1}{2}\ln\left|u^2 +\sqrt{2}u+1\right|+\tan^{-1}\left(\sqrt{2}u+1\right)\right)$

$\therefore\displaystyle\int\sqrt{\cot{x}}\,dx =-\frac{1}{2\sqrt{2}}\left( \ln\left|\cot{x}-\sqrt{2\cot{x}}+1\right| -\ln\left|\cot{x} +\sqrt{2\cot{x}}+1\right|+2\tan^{-1}\left(\sqrt{2\cot{x}}-1\right) +2\tan^{-1}\left(\sqrt{2\cot{x}}+1\right)\right)$ $\square$

ถ้าได้ว่า $\displaystyle\int\sqrt{\cot{x}}\,dx=-\int\frac{2u^2}{u^4+1}\,du$ จะได้ว่ามีค่าเป็น $ -\int\frac{1}{u^4+1}\,du^4$ ซึ่งน่าจะไปต่อง่ายกว่าครับ
ปล. ช่วงนี้รู้สึกจะเข้าช่วง Calculus fever นะครับ เดี๋ยวสอบเสร็จมาเล่นด้วยครับ

[Little Penguin]

ตอบคุณ -SIL-
$\displaystyle \int\frac{1}{u^4+1}\,du^4 =\int\frac{4u^3}{u^4+1}\,du \not=\int\frac{2u^2}{u^4+1}\,du$

วิธีทำข้อ 2 คุณ -InnoXenT- ดูแปลกๆตรง $x=u-e^u$

โจทย์คุณ -InnoXenT-
1.แทน $u=x^2$ แล้วทำ partial fraction ได้คำตอบเป็น $\displaystyle x +\tan^{-1}{\left(e^x\right)} -\frac{1}{2}\ln\left|e^{2x}+1\right|+C$
3.ให้ $\displaystyle \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=u$ ได้คำตอบเป็น $\displaystyle \frac{2}{3}u^3-\frac{1}{2}u^2+C$
4.ทำ By parts 2 ครั้ง ได้คำตอบเป็น $\displaystyle \frac{x}{2}\left(\sin{\left(\ln{x}\right)} -\cos{\left(\ln{x}\right)}\right)+C$
5.ให้ $u=\sqrt{1+\ln{x}}$ ได้คำตอบเป็น $\displaystyle 2\sqrt{1+\ln{x}}+\ln\left|\frac{\sqrt{1+\ln{x}}-1}{\sqrt{1+\ln{x}}+1}\right|+C$
6. ทำ By parts 2 ครั้ง ได้คำตอบเป็น $\dfrac{1}{5}x^5\ln^2{x}-\dfrac{2}{25}x^5\ln{x}+\dfrac{2}{125}x^5+C$