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#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 5

Answer:- $\frac{-1}{7} \cos ^{7} x+\frac{1}{9} \cos ^{9} x+c$

Hint: - We substitute method to solve this integral.

Given:- $\int \sin ^{3} x \cdot \cos ^{6} x d x$

Solution: - Let $I=\int \sin ^{3} x \cdot \cos ^{6} x d x$

The exponent of $\sin x$ is odd, so we substitute
$\operatorname{Cos} x=t \Rightarrow-\operatorname{Sin} x d x=d t$, then
\begin{aligned} &I=\int \sin ^{3} x t^{6} \cdot \frac{d t}{-\operatorname{Sin} x}\quad\quad\quad\quad\quad\left [ t=Cos x \right ]\\ &I=-\int \operatorname{Sin}^{2} x t^{6} d t \end{aligned}
\begin{aligned} &=-\int\left(1-\operatorname{Cos}^{2} x\right) \cdot t^{6} d t \quad\quad\quad\quad\quad\quad\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=1\right] \\ &=-\int t^{6}\left(1-t^{2}\right) d t \end{aligned}
\begin{aligned} &=-\int t^{6}\left(1-t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{6} \cdot t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{8}\right) d t \\ &=-\int t^{6} d t+\int t^{8} d t \end{aligned}
\begin{aligned} &=-\frac{t^{6+1}}{6+1}+\frac{t^{8+1}}{8+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{t^{7}}{7}+\frac{t^{9}}{9}+C \\ &=-\frac{\operatorname{Cos}^{7} x}{7}+\frac{\operatorname{Cos}^{9} x}{9}+C \quad\quad\quad\quad\quad[\because t=\operatorname{Cos} x] \end{aligned}