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  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19 2 question 41 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 41 maths textbook solution.

Answers (1)

Answer   : \frac{1}{8}

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

Given  : \int_{0}^{\frac{\pi}{2}}\sin ^22t.\cos 2tdt

Solution : \int_{0}^{\frac{\pi}{2}}\sin ^22t.\cos 2tdt

\begin{aligned} &\text { Put } \sin 2 t=u \Rightarrow 2 \cos 2 t d t=d u \Rightarrow \cos 2 t d t=\frac{d u}{2}\\ &\text { When } t=\frac{\pi}{4} \text { then } u=1 \text { when } t=0 \text { then } u=0 \end{aligned}

\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cdot \cos 2 t d t \\ &=\int_{0}^{1} u^{3} \frac{d u}{2}=\frac{1}{2} \int_{0}^{1} u^{3} d u \\ &=\frac{1}{2}\left[\frac{u^{3}+1}{3+1}\right]_{0}^{1}=\frac{1}{2}\left[\frac{u^{4}}{4}\right]_{0}^{1} \\ &=\frac{1}{8}\left[u^{4}\right]_{0}{ }^{1} \\ &=\frac{1}{8}\left[1^{4}-0^{4}\right] \\ &=\frac{1}{8} \end{aligned}

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