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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 49.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 49.
 

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 x.\cos x\tan^{-1}(\sin x)dx

Solution:

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

Put  t=\sin x\Rightarrow \cos xdx=dt

when x=0  then t=0  and

when x=\frac{\pi}{2} , then t=1

\therefore \int_{0}^{\pi / 2} 2 \sin x \cdot \cos x \tan ^{-1}(\sin x) d x=\int_{0}^{1} 2 t \cdot \tan ^{-1}(t) d t

Applying integration by parts, method then

\begin{aligned} &=2\left\{\left[\tan ^{-1} t \int t \mathrm{~d} t\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d t}\left(\tan ^{-1} t\right) \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 \end{aligned}

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

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

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