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

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

Answers (1)

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}

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