Question
The value of $$\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{x}{{1 + \sin \,x}}dx} $$ is equal to :
A.
$$\left( {\sqrt 2 - 1} \right)\pi $$
B.
$$\left( {\sqrt 2 + 1} \right)\pi $$
C.
$$\pi $$
D.
none of these
Answer :
$$\left( {\sqrt 2 - 1} \right)\pi $$
Solution :
$$\eqalign{
& I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{x}{{1 + \sin \,x}}} dx \cr
& I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{\left( {\pi - x} \right)}}{{1 + \sin \,x}}} dx \cr
& 2I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{\pi }{{1 + \sin \,x}}} dx \cr
& I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \sin \,x}}} \cr
& I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( {\frac{{1 - \sin \,x}}{{{{\cos }^2}x}}} \right)dx} \cr
& I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( {{{\sec }^2}x - \tan \,x\,\sec \,x} \right)dx} \cr
& I = \frac{\pi }{2}\left[ {\tan \,x - \sec \,x} \right]_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} \cr
& I = \frac{\pi }{2}\left[ {\left( { - 1 + \sqrt 2 } \right) - \left( {1 - \sqrt 2 } \right)} \right] \cr
& I = \pi \left[ {\left( {\sqrt 2 - 1} \right)} \right] \cr} $$
Option A is correct answer.