MCQ
$\int_{ - \,\pi }^{\,\pi } {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $ is
- A${\pi ^2}/4$
- ✓${\pi ^2}$
- C$0$
- D$\pi /2$
$= \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}$.
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($1$) $P(X>Y)$ is
($A$) $\frac{1}{4}$ ($B$) $\frac{5}{12}$ ($C$) $\frac{1}{2}$ ($D$) $\frac{7}{12}$
($2$) $P(X=Y)$ is
($A$). $\frac{11}{36}$ ($B$) $\frac{1}{3}$ ($C$) $\frac{13}{36}$ ($D$) $\frac{1}{2}$
Given the answer quetion ($1$) and ($2$)