#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 46.

Answer :  $\frac{8}{15}$

Hint: Use indefinite formula and the given limits to solve this integral
Given:

$\int_{0}^{\frac{\pi}{2}}\cos ^5x \; dx$

Solution:

\begin{aligned} \int_{0}^{\pi / 2} \cos ^{5} x d x &=\int_{0}^{\pi / 2} \cos ^{4} x \cdot \cos x d x \\ &=\int_{0}^{\pi / 2}\left(\cos ^{2} x\right)^{2} \cos x d x \\ &=\int_{0}^{1 / 2}\left(1-\sin ^{2} x\right)^{2} \cos x d x \quad \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \end{aligned}

Put  $\sin x=t\Rightarrow \cos x \; dx=dt$

when x=0  then t=0

when$x=\frac{\pi}{2}$  then t=1

\begin{aligned} &\therefore \int_{0}^{\pi / 2} \cos ^{5} x d x=\int_{0}^{1}\left(1-t^{2}\right)^{2} d t \\ &\qquad \begin{aligned} & \int_{0}^{1}\left(1+t^{4}-2 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ =& \int_{0}^{1} t^{0} d t+\int_{0}^{1} t^{4} d t-2 \int_{0}^{1} t^{2} d t \end{aligned} \end{aligned}

\begin{aligned} &=\left[\frac{t^{1+1}}{0+1}\right]_{0}^{1}+\left[\frac{t^{4+1}}{4+1}\right]_{0}^{1}-2\left[\frac{t^{2+1}}{2+1}\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \because \int x^{n} d x=\frac{x^{n+1}}{x+1}+c \\ &=[t]_{0}^{1}+\frac{1}{5}\left[t^{5}\right]_{0}^{1}-\frac{2}{3}\left[t^{3}\right]_{0}^{1} \\ &=[1-0]+\frac{1}{5}\left[1^{5}-0^{5}\right]-\frac{2}{3}\left[1^{3}-0^{3}\right] \\ &=1+\frac{1}{5}-\frac{2}{3} \\ &=\frac{15+3-10}{15} \end{aligned}