The value of the integral \int^{\pi}_{-\pi}\frac{\cos^2x}{1+a^x}dx, where a > 0, is

KVPY SX SB Exam 2012 Question Paper

Question : 87 of 120

Marks: +1, -0

The value of the integral

∫↖{π}↙{-π}{\cos^2x}/{1+a^x}dx

, where

a > 0,

Solution:

I=∫↖{π}↙{-π}{\cos^2x}/{1+a^x}dx

.............(1)

[∫↖{b}↙{a}f(x)dx=∫↖{b}↙{a}f(a+b-x)dx]

I=∫↖{π}↙{-π}{\cos^2(-x)}/{1+a^{-x}}dx

I=∫↖{π}↙{-π}{a^x\cos^2 x }/{1+a^{ x}}dx

............(2)
add equation (1) and (2)

2I=∫↖{π}↙{-π}\cos^2x(1/{1+a^x}+a^x/{1+a^x})dx

I=∫↖{π}↙{0}\cos^2xdx

=2∫↖{π\/2}↙{0}\cos^2xdx=π/2

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