#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 9

Answer: - $-\frac{1}{6} \cos ^{6} x+\frac{1}{8} \cos ^{8} x+C$

Hint: - Use substitution method to solve this integral.

Given: -$\int \sin ^{3} x \cdot \cos ^{5} x d x$

Solution: - Let   $I=\int \sin ^{3} x \cdot \cos ^{5} x d x$

Substitute $\cos x=t \Rightarrow-\sin x d x=d t$ then

\begin{aligned} I &=\int\sin ^{3} x t^{5} \cdot \frac{d t}{-\sin x}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left [ \because \cos x=t \right ] \\ &=-\int \sin ^{2} x t^{5} \cdot d t=-\int\left(t-\cos ^{2} x\right) t^{5} d t \quad \quad \quad \quad \quad \quad \quad \left [ \because \sin ^{2}x+\cos ^{2 }x=1 \right ]\\ &=-\int\left(1-t^{2}\right)\left(t^{5}\right) d t=-\int\left(t^{5}-t^{5} t^{2}\right) d t \end{aligned}
\begin{aligned} &=-\int\left(t^{5}-t^{7}\right) d t=-\int t^{5} d t+\int t^{7} d t \\ &=-\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{t^{6}}{6}+\frac{t^{8}}{8}+c=-\frac{\cos ^{6} x}{6}+\frac{\cos ^{8} x}{8}+C \quad\quad\quad\quad\quad\quad\quad[\because \cos x=t] \end{aligned}

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