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 20.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 20.

 

Answers (1)

Answer:\frac{ 2}{3}\tan^{-1}\left (\frac{1}{13} \right )

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

Given: \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx

Solution: I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx

Put \sin x=\frac{2\tan \frac{x}{2} }{1+\tan ^2\frac{x}{2}}

 

I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1}{5+\frac{4\left(2 \tan \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \end{aligned}

=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1+\tan ^{2} x=\sec ^{2} x\right]

Put \tan \frac{x}{2}=t

\sec^ 2\frac{x}{2} .\frac{1}{2}dx=dt

\sec ^2\frac{x}{2} dx=2t

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

\begin{aligned} &l=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \\ &=\int_{0}^{1} \frac{1}{5+5 t^{2}+8 t} 2 d t \\ &=\int_{0}^{1} \frac{1}{5\left(1+t^{2}+\frac{8 t}{5}\right)} 2 d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{1+t^{2}+\frac{8 t}{5}} d t \end{aligned}

\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{t^{2}+2 t \frac{4}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}-\frac{16}{25}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{25-16}{25}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{9}{25}} d t \end{aligned}

\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5}\left[\frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &=\frac{2}{5} \times \frac{5}{3}\left[\tan ^{-1}\left(\frac{\frac{5 t+4}{5}}{\frac{3}{5}}\right)\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5 \times 1+4}{3}\right)-\tan ^{-1}\left(\frac{5 \times 0+4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{3-\frac{4}{3}}{1+3 \times \frac{4}{3}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} A-\tan ^{-1} B=\tan ^{-1}\left(\frac{A-B}{1+A B}\right)\right] \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{\frac{9-4}{3}}{1+4}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5}{3 \times 5}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)\right] \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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question