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  • 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}

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