#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 35

Answer: $\frac{1}{4}\left(\sin ^{-1} x^{2}\right)^{2}+c$

Hint:Use substitution method to solve this integral.

Given$\int \frac{x \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$

Solution:

\begin{aligned} &\text { Let } I=\int \frac{x \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x \\ &\text { Put } \sin ^{-1} x^{2}=t \Rightarrow \frac{1}{\sqrt{1-\left(x^{2}\right)^{2}}} 2 x\; d x=d t \\ &\Rightarrow d x=\frac{\sqrt{1-\left(x^{2}\right)^{2}}}{2 x} d t \Rightarrow d x=\frac{\sqrt{1-x^{4}}}{2 x} d t \text { then } \end{aligned}

\begin{aligned} \Rightarrow I &=\int \frac{x \cdot t}{\sqrt{1-x^{4}}} \frac{\sqrt{1-x^{4}}}{2 x} d t \\ &=\frac{1}{2} \int t d t=\frac{1}{2}\left[\frac{t^{1+1}}{1+1}\right]+c \end{aligned}                                    $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

\begin{aligned} &=\frac{1}{2} \frac{t^{2}}{2}+c=\frac{1}{4} t^{2}+c \\ &=\frac{1}{4}\left(\sin ^{-1} x^{2}\right)^{2}+c \quad \quad\left[\because t=\sin ^{-1} x^{2}\right] \end{aligned}