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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 51.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 51.
 

Answers (1)

Answer :   \pi -2

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

Given:

\int_{0}^{1}(\cos ^{-1}x)^2dx

Solution:

\int_{0}^{1}(\cos ^{-1}x)^2dx

Applying integration by parts, then

\begin{aligned} &=\left[\left(\cos ^{-1} x\right)^{2} \int 1 d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x}\left(\cos ^{-1} x\right)^{2} \int 1 d x\right\} \mathrm{d} x\\ &=\left[\left(\cos ^{-1} x\right)^{2} \cdot x\right]_{0}^{1}-\int_{0}^{1} 2 \cos ^{-1} x \cdot \frac{-1}{\sqrt{1-x^{2}}} \cdot x d x\; \; \; \; \; \; \; \; \; \; \; \; .........(i).\\ &\therefore \int_{0}^{1} 2 \cos ^{-1} x\left(\frac{-1}{\sqrt{1-x^{2}}} \cdot x\right) d x\\ &\text { put } \cos ^{-1} x=t\\ &\Rightarrow \frac{-1}{\sqrt{1-x^{2}}} d x=d \mathrm{t} \end{aligned}

When x=0  then t=\frac{\pi}{2}

 and when x=1  then t=0

therefore, 

\begin{aligned} &-\int_{0}^{1} 2 \cos ^{-1} x \cdot \frac{x}{\sqrt{1-x^{2}}} d x \\ &=-\int_{\pi / 2}^{0} 2 t \cos t d t \\ &=2 \int_{0}^{\pi / 2} t \cos t d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right] \end{aligned}

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

 

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