#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 43.

Answer  : $\frac{3}{4}$

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

Given : $\int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta$

Solution : $I=\int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta$

\begin{aligned} &I==\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos ^{3} 2 \theta} \cdot \sin 2 \theta d \theta \\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sin 2 \theta}{\cos 2 \theta} \cdot \frac{1}{\cos ^{2} 2 \theta} d \theta \\ &I=\int_{0}^{\frac{\pi}{6}} \tan 2 \theta \cdot \sec ^{2} 2 \theta d \theta \end{aligned}

put $\tan 2\theta =t \Rightarrow \sec^22\theta .2d\theta = dt \Rightarrow \sec^22\theta d\theta = \frac{dt}2{}$

when $\theta =0$ then $t=0$ and when $\theta = \frac{\pi}{6}$ then $t=3$

Therefore ,

\begin{aligned} &I=\int_{0}^{\sqrt{3}} t \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{\sqrt{3}} t d t \\ &=\frac{1}{2}\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\sqrt{3}} \end{aligned}

\begin{aligned} &=\frac{1}{2} \cdot \frac{1}{2}\left[t^{2}\right]_{0}^{\sqrt{3}} \\ &=\frac{1}{4}\left[\sqrt{3}^{2}-0^{2}\right] \\ &I=\frac{1}{4} \times 3 \\ &I=\frac{3}{4} \end{aligned}