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

$\Rightarrow dx = - 2\sin \theta \,d\theta ,$ रखने पर,

$dx = - 2\sin \theta \,d\theta $

अब $\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } dx = - 2\int_{\pi /2}^0 {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \sin \theta \,d\theta $

$ = 4\int_0^{\pi /2} {\frac{{\cos (\theta /2)}}{{\sin (\theta /2)}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $

$ = 2\int_0^{\pi /2} {(1 + \cos \theta )\,d\theta } $

$ = 2[\theta + \sin \theta ]_0^{\pi /2} = 2\left[ {\frac{\pi }{2} + 1} \right] = \pi + 2$.