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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 50.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 50.
 

Answers (1)

Answer :   \frac{\pi}{2}-1

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

Given:

\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx

Solution:

\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx=\int_{0}^{\frac{\pi}{2}}2\sin x .\cos x\tan ^{-1}(\sin x)dx

put \sin x= t \Rightarrow \cos xdx=dt
when x=0  then t=0  &

when x=\frac{\pi}{2} then t=1

\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx=\int_{0}^{\frac{\pi}{2}}2\tan ^{-1}(t)dt

 Applying by parts, then

\begin{aligned} &=2\left\{\left[\tan ^{-1} t \int t d t\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d t} \tan ^{-1} t \int t d t\right] d t\right\} \\ &=2\left\{\left[\tan ^{-1} t \cdot \frac{t^{2}}{2}\right]_{0}^{1}-\int_{0}^{1}\left[\frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right]\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{2}\left[t^{2} \tan ^{-1} t\right]_{0}^{1}-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t \\ &=\left[1^{2} \tan ^{-1}(1)-0 \times \tan ^{-1} 0\right]-\int_{0}^{1} \frac{t^{2}+1-1}{1+t^{2}} d t \end{aligned}

\begin{aligned} &=\left[1^{2} \tan ^{-1} 1-0 \times \tan ^{-1} 0\right]-\int_{0}^{1}\left(\frac{1+t^{2}}{1+t^{2}}-\frac{1}{1+t^{2}}\right) d t \\ &=\left[1 \cdot \frac{\pi}{4}-0\right]-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right) d t \\ &=\frac{\pi}{4}-\int_{0}^{1} 1 d t+\int_{0}^{1} \frac{1}{1+t^{2}} d t \end{aligned}

\begin{aligned} &=\frac{\pi}{4}-[t]_{0}^{1}+\left[\tan ^{-1} t\right]_{0}^{1}\ &\left[\because \int 1 d x=x, \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x\right]\\ &=\frac{\pi}{4}-[1-0]+\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\ &\ \left[\therefore \tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right]\\ &=\frac{\pi}{4}-1+\left[\frac{\pi}{4}-0\right]\\ &=\frac{2 \pi}{4}-1\\ &=\frac{\pi}{2}-1 \end{aligned}

 

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