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GATE | GATE 2017 MOCK II | Question 29

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  • Last Updated : 14 Feb, 2018
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The value of the following Integral is:

cos
(A) π
(B) aπ
(C) π/2
(D) 2π


Answer: (C)

Explanation:

Use \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b-x)dx \\ \therefore I=\int_{-\pi }^{\pi }\frac{cos^{2}x}{1+a^{x}}dx, a+b=0 \\ I=\int_{-\pi }^{\pi }\frac{cos^{2}(0-x)}{1+a^{(0-x)}}dx=\int_{-\pi }^{\pi }\frac{a^{x}cos^{2}x}{1+a^{x}}dx \\ 2I=\int_{-\pi }^{\pi }\frac{1+a^{x}}{1+a^{x}}cos^{2}xdx=2\int_{0 }^{\pi }cos^{2}xdx \\ I = 2\int_{0}^{\pi/2 }cos^{2}xdx=2.\frac{1}{2}.\frac{\pi }{2}=\frac{\pi }{2}


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