Question
$$\int_{ - \pi }^\pi {\frac{{2x\left( {1 + \sin \,x} \right)}}{{1 + {{\cos }^2}\,x}}dx} $$ is-
A.
$$\frac{{{\pi ^2}}}{4}$$
B.
$${\pi ^2}$$
C.
zero
D.
$$\frac{\pi }{2}$$
Answer :
$${\pi ^2}$$
Solution :
$$\eqalign{
& \int_{ - \pi }^\pi {\frac{{2x\left( {1 + \sin \,x} \right)}}{{1 + {{\cos }^2}\,x}}dx} \cr
& = \int_{ - \pi }^\pi {\frac{{2x\,dx}}{{1 + {{\cos }^2}\,x}} + 2\int_{ - \pi }^\pi {\frac{{x\,\sin \,x}}{{1 + {{\cos }^2}\,x}}dx} } \cr
& = 0 + 4\int_0^\pi {\frac{{x\,\sin \,x\,dx}}{{1 + {{\cos }^2}\,x}}\,;\,\,\,\,\,\,\,\,\,\left[ {\because \int\limits_{ - a}^a {f\left( x \right)dx = 0} } \right]} \cr} $$
if $$f\left( x \right)$$ is odd $$ = 2\int\limits_0^a {f\left( x \right)dx} $$ if $$f\left( x \right)$$ is even.
$$\eqalign{
& I = 4\int_0^\pi {\frac{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)}}{{1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx \cr
& I = 4\int_0^\pi {\frac{{\left( {\pi - x} \right)\sin \,x}}{{1 + {{\cos }^2}x}}} dx \cr
& \Rightarrow I = 4\pi \int_0^\pi {\frac{{\sin \,x\,dx}}{{1 + {{\cos }^2}x}}} - 4\int {\frac{{x\,\sin \,x\,dx}}{{1 + {{\cos }^2}x}}} \cr
& \Rightarrow 2I = 4\pi \int_0^\pi {\frac{{\sin \,x}}{{1 + {{\cos }^2}x}}dx} \cr
& {\text{Put cos }}x = t\,\, \Rightarrow - \sin \,x\,dx = dt \cr
& \therefore I = - 2\pi \int\limits_1^{ - 1} {\frac{1}{{1 + {t^2}}}dt} = 2\pi \int\limits_{ - 1}^1 {\frac{1}{{1 + {t^2}}}dt} \cr
& = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \cr
& = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right] \cr
& = 2\pi \left[ {\frac{\pi }{4} - \left( {\frac{{ - \pi }}{4}} \right)} \right] \cr
& = 2\pi .\frac{\pi }{2} \cr
& = {\pi ^2} \cr} $$