$=\int_{}^{}\frac{1}{1+\frac{sin^{4}x}{cos^{4}x} } \,dx $
$=\int_{}^{}\frac{cos^4x}{cos^{4}x+sin^{4}x} \,dx $
$=\int_{}^{}\frac{cos^4x}{(cos^{2}x+sin^{2}x)^2-2sin^2xcos^2x} \,dx $
$=\int_{}^{}\frac{2cos^4x}{2-sin^22x} \,dx $
$=\frac{1}{2} \int_{}^{}\frac{(1+cos2x)^2}{2-sin^22x} \,dx $
$=\frac{1}{2} \int_{}^{}\frac{cos^22x+2cos2x+1}{2-sin^22x} \,dx $
$=\frac{1}{2} [\int_{}^{}\frac{cos^22x}{2-sin^22x} \,dx +2\int_{}^{}\frac{cos2x}{2-sin^22x} \,dx +\int_{}^{}\frac{1}{2-sin^22x} \,dx ]$
$=\frac{1}{2} [\int_{}^{}\frac{1-sin^22x}{2-sin^22x} \,dx +2\int_{}^{}\frac{cos2x}{2-sin^22x} \,dx +\int_{}^{}\frac{1}{2-sin^22x} \,dx ]$
$=\frac{1}{2} [{\int_{}^{}(1-\frac{1}{2-sin^22x} )\,dx +}+2\int_{}^{}\frac{cos2x}{2-sin^22x} \,dx +\int_{}^{}\frac{1}{2-sin^22x} \,dx ]$
$=\frac{1}{2} [{\int_{}^{}\,dx +}+2\int_{}^{}\frac{cos2x}{2-sin^22x} \,dx ]$
$=\frac{1}{2} [{\int_{}^{}\,dx +}+\int_{}^{}\frac{1}{2-sin^22x} \,dsin2x ] ---(*)$
ให้ $u=sin2x ,$ และ $u=\sqrt{2} sin\theta $ ; $\frac{-\pi }{2} \leqslant \theta \leqslant \frac{\pi }{2} $
$\therefore du=\sqrt{2} cos\theta d\theta $
$=\frac{1}{2} [{\int_{}^{}\,dx +}+\int_{}^{}\frac{\sqrt{2} cos\theta }{2-2sin^22x} \,d\theta ]$
$=\frac{1}{2} [{\int_{}^{}\,dx +}+\int_{}^{}\frac{\sqrt{2} cos\theta }{2cos^2\theta } \,d\theta ]$
$=\frac{1}{2} [{\int_{}^{}\,dx +}+\frac{\sqrt{2} }{2} \int_{}^{} sec\theta \,d\theta ]$
$=\frac{x}{2}+\frac{\sqrt{2} ln|sec\theta +tan\theta |}{4} +C $
เพราะว่า $sec\theta =\frac{2}{\sqrt{2-u^2} } , tan\theta =\frac{u}{\sqrt{2-u^2} } $
$=\frac{x}{2}+\frac{\sqrt{2} ln|\frac{2-u}{\sqrt{2-u^2 }} |}{4} +C $
$=\frac{x}{2}+\frac{\sqrt{2} ln|\frac{2+u}{2-u} |}{8} +C $
$=\frac{x}{2}+\frac{\sqrt{2} ln|\frac{2+sin2x}{2-sin2x} |}{8} +C $ ##
หมายเหตุ: จากบรรทัด (*) สามารถใช้สูตร $\int_{}^{}\frac{1}{a^2-u^2} \,du = \frac{log|\frac{a+x}{a-x} |}{2a} $ ได้เลยได้คำตอบเท่ากัน
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