Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 60 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 60 maths.

Answers (1)

Answer :   \frac{1}{6}

Hint: Use indefinite integral formula and the limits to solve this integral

Given: \int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx

Solution: \int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx

Dividing numerator and denominator by \cos ^6x

\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{6} x}}{\frac{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}}{\cos ^{6} x}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x}{\cos ^{2} x} \cdot \frac{\cos ^{2} x}{\cos ^{4} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \frac{1}{\cos ^{2} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\frac{\sin ^{3} x}{\cos ^{3} x}+\frac{\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \end{aligned}
Put \tan x =t \Rightarrow \sec^2x\; dx=dt
Now x=0  then t=0 & when x=\frac{\pi}{4} then t=1
\begin{aligned} &\therefore \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} x \cdot \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \\ &=\int_{0}^{1} \frac{t^{2}}{1+t^{6}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{1+\left(t^{3}\right)^{2}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t \end{aligned}

Again putting 1+\frac{t}{3}=u\Rightarrow 3t^2dt=du\Rightarrow t^2dt=\frac{du}{3}
then
\begin{aligned} &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t=\int_{0}^{1} \frac{1}{u^{2}} \frac{d u}{3}=\frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}} d u \\ &=\frac{1}{3} \int_{0}^{1} u^{-2} d u=\frac{1}{3}\left[\frac{u^{-2+1}}{-2+1}\right]_{0}^{1} \\ &=\frac{1}{3}\left[\frac{u^{-1}}{-1}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{u}\right]_{0}^{1} \\ &=-\frac{1}{3}\left[\frac{1}{1+t^{3}}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{1+1}-\frac{1}{1+0}\right] \\ &=-\frac{1}{3}\left[\frac{1}{2}-1\right]=-\frac{1}{3}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{6} \end{aligned}

Posted by

infoexpert24

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question