7.2 definite integral — Maths STD 12 Science — Question

$= \int_{ - \pi }^\pi {\frac{{2x}}{{1 + {{\cos }^2}x}}\,dx} + \int_{ - \pi }^\pi {\frac{{2x\sin x}}{{1 + {{\cos }^2}x}}\,dx} $

==> $I = 0 + \int_{ - \pi }^\pi {\frac{{2x\sin x}}{{1 + {{\cos }^2}x}}\,dx} $

$\left[ {\left. {\begin{array}{*{20}{c}}{\int_{ - a}^a {f(x)dx = 2\int_0^a {f(x)\,dx} ,} }&{{\rm{if }}f( - x) = f(x)}\\{\,\, = 0,}&{{\rm{if }}f( - x) = - f(x)}\end{array}} \right]} \right.$

$ \Rightarrow I = 2\int_0^\pi {\frac{{2x\,\,\sin x}}{{1 + {{\cos }^2}x}}} \,dx$ 

$ \Rightarrow I = 4\int_0^\pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}\,dx} $........$(i)$

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

$\left( \because \int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \right)$

Adding $(i)$ and $(ii),$ we get

$ \Rightarrow 2I = 4\int_0^\pi {\frac{{\pi \,\,\sin x}}{{1 + {{\cos }^2}x}}} \,dx$ 

$ \Rightarrow I = 2\pi \int_0^\pi {\frac{{\,\,\sin x}}{{1 + {{\cos }^2}x}}dx} $

Put $\cos x = t\,\,$==> $ - \sin x\,\,dx = dt$

$ \Rightarrow I = 2\pi \int_1^{ - 1} {\frac{{ - dt}}{{1 + {t^2}}}} $

$ \Rightarrow I = - 2\pi \,[{\tan ^{ - 1}}t]\,_1^{ - 1}$

$ \Rightarrow I = - 2\pi \left( {\frac{{ - \pi }}{4} - \frac{\pi }{4}} \right) = {\pi ^2}$.