# integral x/(1+cos^2x) from 0 to pi

Jitender Singh IIT Delhi
9 years ago
Ans:
$I = \int_{0}^{\pi }\frac{x}{1+cos^{2}x}dx$
$I = \int_{0}^{a }f(x)dx = \int_{0}^{a}f(a-x)dx$
$I = \int_{0}^{\pi }\frac{\pi -x}{1+cos^{2}(\pi -x)}dx$
$I = \int_{0}^{\pi }\frac{\pi -x}{1+cos^{2}(x)}dx$
$I = \int_{0}^{\pi }\frac{\pi}{1+cos^{2}(x)}dx-\int_{0}^{\pi }\frac{x}{1+cos^{2}x}dx$
$I = \int_{0}^{\pi }\frac{\pi}{1+cos^{2}(x)}dx-I$
$2I = \int_{0}^{\pi }\frac{\pi}{1+cos^{2}(x)}dx$
$I = \frac{\pi }{2}\int_{0}^{\pi }\frac{1}{1+cos^{2}(x)}dx$
$I = \frac{\pi }{2}\int_{0}^{\pi }\frac{sec^{2}(x)}{1+cos^{2}(x)}.\frac{1}{sec^{2}(x)}dx$
$I = \frac{\pi }{2}\int_{0}^{\pi }\frac{sec^{2}(x)}{1+sec^{2}(x)}dx$
$I = \frac{\pi }{2}\int_{0}^{\pi }\frac{sec^{2}(x)}{tan^{2}(x)+2}dx$
Since sec(x) is not continuous at pi/2, we need to divide the integral
$I = \frac{\pi }{2}[\int_{0}^{\pi/2 }\frac{sec^{2}(x)}{tan^{2}(x)+2}dx + \int_{\pi /2}^{\pi }\frac{sec^{2}(x)}{tan^{2}(x)+2}dx]$
$tan(x) = t$
$sec^{2}(x)dx = dt$
$x = 0,\pi \rightarrow t = 0$
$x =\frac{\pi }{2}^{-} \rightarrow t = \infty$
$x =\frac{\pi }{2}^{+} \rightarrow t = -\infty$
$I = \frac{\pi }{2}[\int_{0}^{\infty }\frac{1}{t^{2}+2}dt + \int_{-\infty }^{0}\frac{1}{t^{2}+2}dt]$
$I = \frac{\pi }{2}[2\int_{0}^{\infty }\frac{1}{t^{2}+2}dt]$
$I = \frac{\pi }{2}[\int_{0}^{\infty }\frac{1}{(\frac{t}{\sqrt{2}})^{2}+1}dt]$
$I = \frac{\pi }{2}[\frac{tan^{-1}(\frac{t}{\sqrt{2}})}{\frac{1}{\sqrt{2}}}]_{0}^{\infty }$
$I = \frac{\pi }{2}[\frac{\pi }{\sqrt{2}}]$
$I = \frac{\pi ^{2}}{2\sqrt{2}}$
Thanks & Regards
Jitender Singh
IIT Delhi
3 years ago
Dear student,