#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 7

Answer: -  $-\frac{1}{8} \operatorname{Cos}^{4} x^{2}+C$

Hint: - Use substitution method to solve this integral.

Given: -   $\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$

Solution: - Let   $I=\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$

Substitute, $\operatorname{Cos} x^{2}=t \Rightarrow-\operatorname{Sin} x^{2} 2 x d x=d t \text { , then }$
\begin{aligned} I &=\int t^{3} x \operatorname{Sin} x^{2} \cdot \frac{d t}{-\left(\sin x^{2}\right)(x)^{2}}\quad\quad\quad\quad &\left[\because \operatorname{Cos} x^{2}=t\right] \\ &=-\frac{1}{2} \int t^{3} d t=\frac{-1}{2} \cdot \frac{t^{3+1}}{3+1}+C &\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
\begin{aligned} &=\frac{-1}{2} \cdot \frac{t^{4}}{4}+C=\frac{-1}{8} t^{4}+C \\ &=\frac{-1}{8} \cos ^{4} x^{2}+C \quad \quad \quad \quad \quad \quad\left[\because t=\cos x^{2}\right] \end{aligned}