Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 11.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 11.

Answers (1)

Answer: \frac{64}{231}

Hint: We use indefinite integral formula then put limits to solve this integral.

Given:\int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi

Solution: \int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi

Put \sin \phi=t

\cos \phi d\phi =dt

d\phi =\frac{dt}{cos\phi}

When \phi =0  then t=0  and when \phi =\frac{\pi}{2}  then t=1

\begin{aligned} &=\int_{0}^{1} \sqrt{t} \cos ^{5} \phi \frac{d t}{\cos \phi} \\ &=\int_{0}^{1} \sqrt{t} \cos ^{4} \phi d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(\cos ^{2} \phi\right)^{2} d t \end{aligned}

\begin{aligned} &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-\sin ^{2} \phi\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-t^{2}\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t \end{aligned}

\begin{aligned} &=\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{4+\frac{1}{2}}-2 t^{2+\frac{1}{2}}\right) d t \\ &=\int_{0}^{1} t^{\frac{1}{2}} d t+\int_{0}^{1} t^{\frac{9}{2}} d t-2 \int_{0}^{1} t^{\frac{5}{2}} d t \end{aligned}

\begin{aligned} &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2 t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1} \\ &=\left[\frac{t^{\frac{3}{2}}}{3}\right]_{0}^{1}+\left[\frac{t^{\frac{11}{2}}}{\frac{11}{2}}\right]_{0}^{1}-2\left[\frac{t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{11}\left[t^{\frac{11}{2}}\right]_{0}^{1}-2 \times \frac{2}{7}\left[t^{\frac{7}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]+\frac{2}{11}\left[1^{\frac{11}{2}}-0^{\frac{11}{2}}\right]-\frac{4}{7}\left[1^{\frac{7}{2}}-0^{\frac{7}{2}}\right] \\ &=\frac{2}{3}[1-0]+\frac{2}{11}[1-0]-\frac{4}{7}[1-0] \end{aligned}

=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}

=\frac{154+42-132}{231}

=\frac{196-132}{231}

=\frac{64}{231}

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 Date Sheet 2027 Live: Class 10, 12 exam dates soon on cbse.gov.in; private candidate form, LOC submission
anam-khan

CBSE Date Sheet 2027 LIVE: 10th, 12th board exam schedule soon; simple papers, passing marks
anam-khan

CBSE Date Sheet 2027 LIVE: Class 10, 12 board exam schedule soon; passing marks, paper pattern

Latest Question