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Grade 11Integral Calculus

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

Shadow

12 Years agoGrade 11

2 Answers

Jitender Singh

12 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

askIITians Faculty

Kushagra Madhukar

6 Years ago

Dear student,

Please find the solution to your problem.

Now, sec(x) is not continuous at x = π/2, we have to split the integral at the point of discontinuity.

Hence,

Thanks and regards,

Kushagra

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