#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 8

Answer: -$-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{1}{7} \cos ^{7} x+C$

Hint: - Use substitution method to solve this integral.

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

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

Re-writing,
\begin{aligned} &I=\int \sin ^{6} x \cdot \sin x d x \\ &\Rightarrow I=\int\left(\sin ^{2} x\right)^{3} \sin x d x \\ &\Rightarrow I=\int\left(1-\cos ^{2} x\right) \sin x d x \quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=I\right] \end{aligned}
\begin{aligned} &\Rightarrow I=\int\left[1-\cos ^{6} x-3 \cos ^{2} x+3 \cos ^{4} x\right] \sin x d x \\ &\Rightarrow I=\int\left[\sin x-\cos ^{6} x \sin x-3 \cos ^{2} x \sin x+3 \cos ^{4} x \sin x\right] \\ &\Rightarrow I=\int \sin d x-\int \cos ^{6} \cdot \sin x d x-3 \int \cos ^{2} x \sin x d x+3 \int \cos ^{4} x \sin x d x \end{aligned}

Substitute,$\cos x=t \Rightarrow-\sin x d x=d t$ in 2nd, 3rd and 4th integral, then

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