Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 48.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 48.
 

Answers (1)

Answer :  \frac{8}{3}

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

Given:

\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx

Solution:

\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx

put (\cos x)=t

\Rightarrow -\sin x\; dx=dt

\Rightarrow dx=dt - \sin x
 When x=0  then t=1  

when x=\pi  then t=-1

\begin{aligned} &\therefore \int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} d x \\ &=\int_{1}^{-1} \sin ^{3} x(1+2 t)(1+t)^{2} \frac{d t}{-\sin x} \\ &=-\int_{+1}^{-1} \sin ^{2} x(1+2 t)\left(1+t^{2}+2 t\right) d t \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 \alpha b\right] \end{aligned}

\begin{aligned} &=-\int_{+1}^{-1}\left(1-\cos ^{2} x\right)\left(1+t^{2}+2 t+2 t+2 t^{3}+4 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \\ &=-\int_{+1}^{-1}\left(1-t^{2}\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}-1\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}+5 t^{4}+4 t^{3}+2 t^{5}-1-5 t^{2}-4 t-2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(2 t^{5}+2 t^{3}+5 t^{4}-4 t^{2}-4 t-1\right) d t \end{aligned}

\begin{aligned} &=2 \int_{1}^{-1} t^{5} d t+2 \int_{1}^{-1} t^{3} d t+5 \int_{1}^{-1} t^{4} d t-4 \int_{.1}^{-1} t^{2} d t-4 \int_{-1}^{-1} t d t-\int_{1}^{-1} d t \\ &=2\left[\left.\frac{t^{5+1}}{5+1}\right|_{1} ^{-1}+2\left[\frac{t^{3+1}}{3+1}\right]_{1}^{-1}+5\left[\left.\frac{t^{4+1}}{4+1}\right|_{1} ^{-1}-4\left[\left.\frac{t^{2+1}}{2+1}\right|_{1} ^{-1}-4\left[\frac{t^{1+1}}{1+1}\right]_{1}^{-1}-\left[\left.t\right|_{1} ^{-1}\right.\right.\right.\right. \\ &=\frac{2}{6}\left[t^{6}\right]_{1}^{-1}+\frac{2}{4}\left[t^{4}\right]_{1}^{-1}+\frac{5}{5}\left[t^{5}\right]_{1}^{-1}-\frac{4}{3}\left[t^{3}\right]_{1}^{-1}-4\left[\frac{t^{2}}{2}\right]_{1}^{-1}-[t]_{1}^{-1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \end{aligned}

\begin{aligned} &=\frac{2}{6}\left[(-1)^{6}-1^{6}\right]+\frac{2}{4}\left[(-1)^{4}-1^{4}\right]+\left[(-1)^{5}-1^{5}\right]-\frac{4}{3}\left[(-1)^{3}-1\right]-\frac{4}{2}\left[(-1)^{2}-1\right]-[-1-1] \\ &=\frac{1}{3}[1-1]+\frac{1}{2}[1-1]+[-1-1]-\frac{4}{3}[-1-1]-2[1-1]-(-2) \\ &=0+0+(-2)-\frac{4}{3}(-2)-2 \times 0+2 \\ &=\frac{8}{3} \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

CBSE Board 12th Exams 2026 LIVE: Class 12 political science exam 'moderate' in difficulty; simple MCQs
anam-khan

CBSE Class 12 Economics Analysis: Students find exam moderately difficult; questions mostly application-based
anam-khan

CBSE Board Exam Analysis 2026 LIVE: Class 12 economics paper 'moderate to difficult level'; updates

Latest Question