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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
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
askIITians Faculty
Kushagra Madhukar
askIITians Faculty 628 Points
3 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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