Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Evaluate the definite integrals in Exercises 25 to 33. 26

Evaluate the definite integrals in Exercises 25 to 33.

    Q26.  \int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

Answers (1)

best_answer

\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let's divide both the numerator and denominator by cos^4x

=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}

=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}

Now let's change the variable

\\t=tan^2x \\dt=2tanxsec^2xdx

the limits will also change since the variable is changing 

when\:x=0,t=tan^20=0

when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1

So, the integration becomes:

I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}

I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1

I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]

I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0

I=\frac{\pi}{8}

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question