#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 44.

Answer  : $\frac{\pi}{3}$

Hint   : use indefinite formula and the limit to solve this integral

Given   : $\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$

Solution  :  $\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$

put $x^{\frac{3}{2}}=t \Rightarrow \frac{3}{2} x^{\frac{3}{2}-1} d x=d t \Rightarrow \frac{3}{2} x^{\frac{1}{2}} d x=d t \Rightarrow d x=\frac{2}{3 \sqrt{x}} d t$

when x=0 then t=0 when $x=(\Pi)^{\frac{2}{3}}$ then $t=\Pi$

Therefore

$\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$

\begin{aligned} &=\int_{0}^{\pi} \sqrt{x} \cos ^{2} t \frac{2}{3 \sqrt{x}} d t \\ &=\frac{2}{3} \int_{0}^{\pi} \cos ^{2} t d t \\ &=\frac{2}{3} \int_{0}^{\pi}\left(\frac{1+\cos 2 t}{2}\right) d t \\ &=\frac{1}{3} \int_{0}^{\pi}(1+\cos 2 t) d t \\ &=\frac{1}{3}\left[\int_{0}^{\pi} 1 d t\right]+\frac{1}{3}\left[\int_{0}^{\pi} \cos 2 t d t\right] \\ &=\frac{1}{3}[t]_{0}{ }^{\pi}+\frac{1}{3}\left[\frac{\sin 2 t}{2}\right]_{0}^{\pi} \\ &=\frac{1}{3} \Pi-0+\frac{1}{6}[\sin 2 \pi-\sin 2 \times 0] \end{aligned}

$=\frac{1}{3}\Pi+\frac{1}{6}(0-0)$

$=\frac{\Pi}{3}$