อ้างอิง:
ข้อความเดิมเขียนโดยคุณ -InnoXenT-
ขุดจากหน้า 10 
Evaluate
$$\int_{0}^{\frac{\pi}{8}} \frac{\cos{x}}{\cos{(x-\frac{\pi}{8})}} \, dx$$ |
ให้ $ u=x-\frac{\pi}{8} $
จะได้ \[
\int\limits_0^{\frac{\pi }{8}} {\frac{{\cos x}}{{\cos \left( {x - \frac{\pi }{8}} \right)}}dx = } \int\limits_{ - \frac{\pi }{8}}^0 {\frac{{\cos \left( {u + \frac{\pi }{8}} \right)}}{{\cos u}}du = } \int\limits_{ - \frac{\pi }{8}}^0 {\sec u\cos \left( {u + \frac{\pi }{8}} \right)du = \int\limits_{ - \frac{\pi }{8}}^0 {\sec u\left( {\cos u\cos \frac{\pi }{8} - \sin u\sin \frac{\pi }{8}} \right)du} }
\]
\[
= \int\limits_{ - \frac{\pi }{8}}^0 {\left( {\cos \frac{\pi }{8} - \tan u\sin \frac{\pi }{8}} \right)du} = \frac{{\pi \sqrt {2 + \sqrt 2 } }}{{16}} - \frac{{\sqrt {2 - \sqrt 2 } }}{2}\ln \left( {\frac{{\sqrt {2 + \sqrt 2 } }}{2}} \right)
\]