#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 31 Maths Textbook Solution.

Answer: $x\sin ^{-1}\left ( \sqrt{x} \right )+\frac{1}{2}\left [ \sqrt{1-x}\sqrt{x}+\sin ^{-1} \sqrt{1-x}\right ]+c$

Given:$\int \sin ^{-1}\sqrt{xdx}$

Hint: $u\int vdx-\int \left ( \frac{d}{dx} u\int vdx\right )dx$

Solution:

\begin{aligned} &u=\int \sin ^{-1} \sqrt{x} d x \\ &=\int \sin ^{-1} \sqrt{x} \cdot 1 d x \\ &=\sin ^{-1} \sqrt{x} \int 1 d x-\int\left(\frac{d}{d x}\left(\sin ^{-1} \sqrt{x}\right) \int 1 d x\right) d x \end{aligned}

\begin{aligned} &{\left[\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}\right]} \\ &=x \sin ^{-1}(\sqrt{x})-\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \times \frac{1}{2 \sqrt{x}} x d x \\ &=x \sin ^{-1}(\sqrt{x})-\int \frac{\sqrt{x}}{2 \sqrt{1-(\sqrt{x})^{2}}} d x \end{aligned}

Now,

$\sqrt{1-\left ( \sqrt{x} \right )^{2}}=t$

$\Rightarrow \sqrt{1-x}=t$

$\Rightarrow 1-x=t^{2}$

$\Rightarrow 1-t^{2}=x$

$\Rightarrow \sqrt{x}=\sqrt{1-t^{2}}$

Again

$\Rightarrow 1-x=t^{2}$

$-dx=2tdt$

$dx=-2tdt$

$\left [ \int \sqrt{a^{2}-x^{2}}dx =\frac{1}{2} \left [ x\sqrt{a^{2}-x^{2}} a^{2}\sin ^{-1}\frac{x}{a}\right ]\right ]$

\begin{aligned} &=x \sin ^{-1}(\sqrt{x})-\frac{1}{2} \int \frac{\sqrt{1-t^{2}}}{t}(-2 t d t) \\ &=x \sin ^{-1}(\sqrt{x})+\frac{1}{2} \times 2 \int \sqrt{1-t^{2}} d t \\ &=x \sin ^{-1}(\sqrt{x})+\frac{1}{2}\left[t \sqrt{1-t^{2}}+1 \sin ^{-1} t\right]+c \\ &=x \sin ^{-1}(\sqrt{x})+\frac{1}{2}\left[\sqrt{1-x} \sqrt{x}+\sin ^{-1} \sqrt{1-x}\right]+c \end{aligned}