#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 18 Maths Textbook Solution.

Answer: $\frac{2}{3}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{3}xdx$

Solution:$\int_{0}^{\frac{\pi}{2}} \cos ^{3} x d x=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos 3 x+3 \cos x}{4}\right) d x \quad\left[\begin{array}{l} \cos 3 x=4 \cos ^{3} x-3 \cos x \\ \Rightarrow \cos 3 x+3 \cos x=4 \cos ^{3} x \\ \Rightarrow \cos ^{3} x=\frac{\cos 3 x+3 \cos x}{4} \end{array}\right]$

\begin{aligned} &=\int_{0}^{\frac{\pi}{3}}\left(\frac{\cos 3 x}{4}+\frac{3}{4} \cos x\right) d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos 3 x+\frac{3}{4} \int_{0}^{\frac{\pi}{2}} \cos x d x \\ &=\frac{1}{4}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}[\sin x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{4 \times 3}\left[\sin 3 \times \frac{\pi}{2}-\sin 3 \times 0\right]+\frac{3}{4}\left[\sin \frac{\pi}{2}-\sin 0\right] \quad\left[\begin{array}{l} \left.\sin \frac{3 \pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\right] \\ =-\sin \frac{\pi}{2}=-1 \\ \sin \frac{\pi}{2}=1 \end{array}\right. \end{aligned}

$=\frac{1}{12}\left [ -1-0 \right ]+\frac{3}{4}\left [ 1-0 \right ]$

$=\frac{-1}{12}+\frac{3}{4}=\frac{-1+9}{12}$

$=\frac{8}{12}=\frac{2}{3}$