0

Guest

Courses New

Can you please give me the step wise solution....Thank you

Can you please give me the step wise solution....Thank you

Question Image
Grade:12

1 Answers

BALAJI ANDALAMALA
askIITians Faculty 78 Points
8 years ago
From the properties of definite integrals ,we have

\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx

Let\,\,I=\int_{-\pi}^{\pi}\frac{sin^2x}{1+a^x}dx….....................(1)
From the above property , we get

I=\int_{-\pi}^{\pi}\frac{sin^2x}{1+a^{-x}}dx =\int_{-\pi}^{\pi}\frac{a^xsin^2x}{1+a^{x}}dx….........................(2)

Adding (1) and (2) we get

I+I=2I=\int_{-\pi}^{\pi}\frac{(1+a^x)sin^2x}{1+a^{x}}dx =\int_{-\pi}^{\pi}sin^2x dx
=\int_{-\pi}^{\pi}\frac{1-cos2x}{2}dx = \frac{1}{2}\left [ x-\frac{sin2x}{2} \right ]_{-\pi}^{\pi}
\Rightarrow 2I =\pi
\therefore I =\frac{\pi}{2}

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More