STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$ = \int_0^{\pi /4} {\frac{{\cos x\,dx}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} $

$ = \frac{1}{3}\int_0^{1/\sqrt 2 } {\left( {\frac{1}{{1 - {t^2}}} + \frac{2}{{1 + 2{t^2}}}} \right)} \,dt$

आंशिक भिन्नों से, जहाँ $t = \sin x$

$ = \frac{1}{3}\left[ {\frac{1}{{2.1}}\log \frac{{1 + t}}{{1 - t}} + \frac{2}{{\sqrt 2 }}{{\tan }^{ - 1}}t\sqrt 2 } \right]_0^{1/\sqrt 2 }$

$ = \frac{1}{3}\left[ {\frac{1}{2}\log \frac{{(\sqrt 2 + 1)}}{{(\sqrt 2 - 1)}} + \sqrt 2 {{\tan }^{ - 1}}1} \right]$

$ = \frac{1}{3}\left[ {\frac{1}{2}\log {{(\sqrt 2 + 1)}^2} + \sqrt 2 .\frac{\pi }{4}} \right]$

$ = \frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$.