The value of $\int_\limits{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0$ is

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of

\int\limits_{{-\pi}}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0

View answer and explanation

Correct option: C

\frac{\pi}{2}

Option C is correct because by using the property

\int_a^b f(x) dx = \int_a^b f(a+b-x) dx

and adding the two integrals, we can simplify the integral to

\int_{-\pi}^{\pi} \cos^2 x dx

which equals

\pi

, and hence I =

\frac{\pi}{2}

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