#### Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 3

Hint: Use substitution method to solve this integral

Given:  $\int \cos ^{5} x d x$

Solution: Let   $I=\int \cos ^{5} x d x$

Re-writing,   $I=\int \cos ^{5} x d x$
\begin{aligned} I &=\int \cos ^{3} x \cos ^{2} x d x \\ &=\int \cos ^{3} x\left(1-\sin ^{2} x\right) d x\quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}
\begin{aligned} &=\int\left(\cos ^{3} x-\cos ^{3} x \sin ^{2} x\right) d x \\ &=\int \cos ^{3} x d x-\int \cos ^{3} x \sin ^{2} x d x \\ &=\int \cos ^{2} x \cos x d x-\int \cos ^{2} x \cos x \sin ^{2} x d x \end{aligned}
\begin{aligned} &=\int \cos x\left(1-\sin ^{2} x\right) d x-\int\left(1-\sin ^{2} x\right) \sin ^{2} x \cos x d x \quad \quad \quad \quad \quad \quad \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &=\int \cos x d x-\int \cos x \sin ^{2} x d x-\int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \\ &=\int \cos x d x-2 \int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \end{aligned}

Substitute $\sin x=t\Rightarrow \cos xdx=dt$ in second and third integral, then
$I =\int \cos x d x-2 \int t^{2} d t+\int t^{4} d t \quad\quad\quad\quad\quad\quad\left [ \because \int \cos xdx=\sin x+c \right ]$
\begin{aligned} &=\sin x-2 \frac{t^{2+1}}{2+1}+\frac{t^{4+1}}{4+1}+C \\ &=\sin x-2 \frac{t^{3}}{3}+\frac{t^{5}}{5}+C \\ &=\sin x-2 \frac{\sin ^{3} x}{3}+\frac{\sin ^{5} x}{5}+C \end{aligned}