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

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

Grade:11

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 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
askIITians Faculty 629 Points
one year 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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