A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{{{\pi }^{2}}}{2}\]
D) \[\frac{{{\pi }^{2}}}{4}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi }{\frac{x}{2\,\text{cosec x - sin x}}}\,\,dx\] \[\Rightarrow \] \[I=\int{\frac{x}{\frac{2}{\sin x}-\sin x}}\,\,dx\] \[\Rightarrow \] \[I=\int_{0}^{2}{\frac{x\sin x}{2-{{\sin }^{2}}x}}\,dx\] ?. (i) \[\Rightarrow \] \[I=\int_{0}^{\pi }{\frac{(\pi -x)\sin (\pi -x)}{2-{{\sin }^{2}}(-x)}}\,dx\] \[\left[ \because \,\,\int_{0}^{a}{f(x)\,\,dx}=\int_{0}^{a}{f(a-x)\,dx} \right]\] \[\Rightarrow \] \[I=\int_{0}^{\pi }{\frac{(\pi -x)\sin x}{2-{{\sin }^{2}}x}}\,\,dx\] ?.. (ii) On adding Eqs. (i) and (ii), we get \[\Rightarrow \]\[2I=\int_{0}^{\pi }{\frac{x\,\sin x}{2-{{\sin }^{2}}x}}\,dx+\int_{0}^{\pi }{\frac{(\pi -x)\sin x}{2-{{\sin }^{2}}x}\,\,dx}\] \[\Rightarrow \]\[2I=\int_{0}^{\pi }{\left[ \frac{x\,\sin x}{2-{{\sin }^{2}}x}+\frac{\pi \sin x-x\sin x}{2-{{\sin }^{2}}x} \right]}dx\] \[\Rightarrow \] \[2I=\int_{0}^{\pi }{\frac{\pi \sin x}{2-{{\sin }^{2}}x}\,\,dx}\] \[\Rightarrow \]\[2I=\int_{0}^{\pi }{\frac{\pi \sin x}{2-(1-{{\cos }^{2}}x)}\,dx}\] \[\Rightarrow \]\[2I=\pi \int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}\,dx}\] \[\Rightarrow \]\[I=\frac{\pi }{2}\int_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}\,dx}\] Put \[\cos \,\,x=t\] \[\Rightarrow \] \[-\sin x\,\,dx=dt\] or \[\sin xdx=-dt\] Also, when \[x=0,\] then \[t=-1\] when, \[x=\pi ,\] then \[t=-1\] Then, \[I=\frac{\pi }{2}\left[ \int_{1}^{-1}{\frac{(-dt)}{1+{{t}^{2}}}} \right]\] \[=\frac{\pi }{2}\left[ \int_{-1}^{1}{\frac{dt}{1+{{t}^{2}}}} \right]\,\,\,\left[ \because \,\,\int_{b}^{a}{f(x)=-\int_{a}^{b}{f(x)\,\,dx}} \right]\] \[=\frac{\pi }{2}\left[ 2\int_{0}^{1}{\frac{dt}{1+{{t}^{2}}}} \right]\] \[\left[ \because \,\,\int_{-a}^{a}{f(x)=dx=2\int_{0}^{a}{f(x),\,if\,f(-x)=f(x)}} \right]\] \[=\frac{\pi }{2}[2\{{{\tan }^{-1}}t\}_{0}^{1}]\] \[=\frac{\pi }{2}[2\{{{\tan }^{-1}}1-{{\tan }^{-1}}0)\}]\] \[=\frac{\pi }{2}\left[ 2\left\{ \frac{\pi }{4}-0 \right\} \right]\] \[=\frac{\pi }{2}\times 2\times \frac{\pi }{4}\] \[\Rightarrow \] \[I=\frac{{{\pi }^{2}}}{4}\]
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