Evaluate:
          \int_{0}^{\pi }\frac{x\sin x}{1+\cos ^{2}x}dx

 

 

 

 
 
 
 
 

Answers (1)

I= \int_{0}^{\pi }\frac{x\sin x}{1+\cos ^{2}x}dx
                 \because \int_{a}^{b}f(x)= \int_{a}^{b}f(a+b -x)
I= \int_{0}^{\pi }\frac{\left ( \pi -x \right )\sin \left ( \pi -x \right )}{1+\cos ^{2}\left ( \pi -x \right )}dx
I= \int_{0}^{\pi }\frac{\left ( \pi -x \right )\sin x}{1+\cos ^{2}x}dx
I= \int_{0}^{\pi }\frac{\pi \sin x}{1+\cos ^{2}x}dx-\int_{0}^{\pi }\frac{x\sin x}{1+\cos ^{2}x}dx
I= \int_{0}^{\pi }\frac{\pi \sin x}{1+\cos ^{2}x}dx-I
I+I= \int_{0}^{\pi }\frac{\pi \sin x}{1+\cos ^{2}x}dx
\cos x= t
-\sin xdx= dt
= 2I= \int_{0}^{\pi }\frac{-\pi dt}{1+t^{2}}             = I= \frac{-\pi }{2}\int_{0}^{\pi }\frac{dt}{1+t^{2}}
\Rightarrow I= \frac{-\pi }{2}\left [ +\tan^{-1}\left ( t \right ) \right ]^{\pi }_{0}
\Rightarrow I= \frac{-\pi }{2}\left [ \tan^{-1}\left ( \cos x \right ) \right ]^{\pi }_{0}
I= \frac{-\pi }{2}\left [ \tan^{-1}\left ( \cos \pi \right )-\tan^{-1}\cos 0 \right ]
I= \frac{-\pi }{2}\left [ \tan^{-1}\left ( -1 \right )-\tan^{-1}\left ( 1 \right ) \right ]
I= \frac{-\pi }{2}\left [ \frac{-\pi }{4}-\frac{\pi }{4} \right ]
I= \frac{-\pi }{2}\left ( \frac{-2\pi }{4} \right )
I\Rightarrow \frac{\pi ^{2}}{4}


 

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