#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 6

Answer:-  $\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+c$

Hint: - Use substitute method to solve this integral.

Given:- $\int \operatorname{Cos}^{7} x d x$

Solution: - Let  $I=\int \operatorname{Cos}^{7} x d x$

Re-writing,
\begin{aligned} &I=\int \operatorname{Cos}^{6} x \cdot \operatorname{Cos} x d x \\ &\Rightarrow I=\int\left(\operatorname{Cos}^{2} x\right)^{3} \cdot \operatorname{Cos} d x \end{aligned}
$\begin{array}{ll} \Rightarrow I=\int\left(1-\operatorname{Sin}^{2} x\right)^{3} \cdot \operatorname{Cos} d x &\quad\quad\quad\quad\quad {\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=I\right]} \\\\ \Rightarrow I=\int\left[1-\operatorname{Sin}^{6} x-3 \operatorname{Sin}^{2} x+3 \operatorname{Sin}^{4} x\right] \cdot \operatorname{Cos} d x &\quad\quad\quad\quad {\left[\because(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{b}\right]} \end{array}$
\begin{aligned} &\Rightarrow I=\int\left(\operatorname{Cos} x-\operatorname{Sin}^{6} x \cdot \operatorname{Cos} x-3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} x+3 \operatorname{Sin}^{4} x \cdot \operatorname{Cos} x\right) d x \\ &\Rightarrow I=\int \operatorname{Cos} d x-\int \operatorname{Sin}^{6} x \cdot \operatorname{Cos} d x-\int 3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} d x+3 \int \sin ^{4} x \cdot \cos x d x \end{aligned}
Substitute $\operatorname{Sin} x=t \Rightarrow \operatorname{Cos} x d x=d t$ in 2nd, 3rd and 4th integral, then

\begin{aligned} &I=\int \operatorname{Cos} d x-\int t^{6} d t-3 \int t^{2} d t+3 \int t^{4} \cdot d t \\ &=\operatorname{Sin} x-\frac{t^{6}+1}{6+1}-3 \frac{t^{2+1}}{2+1}+3 \frac{t^{4+1}}{4+1}+C \quad \quad \quad \quad \quad \quad \quad\left[\because \int \operatorname{Cos} x d x=\operatorname{Sin} x+c \& \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
\begin{aligned} &=\operatorname{Sin} x-\frac{t^{7}}{7}-3 \frac{t^{3}}{3}+3 \cdot \frac{t^{5}}{5}+C \\ &=\operatorname{Sin} x-\frac{\operatorname{Sin}^{7} x}{7}-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x+C \\ &=\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+C \quad\quad\quad\quad\quad\quad\quad[\because \operatorname{Sin} x=t] \end{aligned}