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

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}