Question 116 of 246MediumMCQ
\[\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx\] is
Solutions · 1
Official solution
\[\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx\]
\[ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx\]
\[ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;\]
\[\left[ \, \right.\] as \[\int\limits_{ - a}^a {f\left( x \right)} dx = 0\] \[\left. \, \right]\]
if f(x) is odd
\[ = 2\int\limits_0^a {f\left( x \right)} dx\] if f(x) is even.
\[I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx\]
\[I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} \]
\[ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} \]
\[ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx\]
put \[\cos x = t \Rightarrow - \sin xdx = dt\]
\[\therefore\] \[I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt\]
\[ = 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt\]
\[ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1\]
\[ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]\]
\[ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}\]
Answer the question to unlock