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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 27

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 27

Answers (1)

Answer: \frac{1}{2}

Hint: you must know the rule of integration

Given: \int_{0}^{\frac{\pi}{4}} \sin 2 x \; d x

Solution:  \int_{0}^{\frac{\pi}{4}} \sin 2 x \; d x

\begin{aligned} &=\frac{1}{2}(-\cos 2 x)_{0}^{\frac{\pi}{4}} \\\\ &=-\frac{1}{2}(\cos 2 x)_{0}^{\frac{\pi}{4}} \end{aligned}

\begin{aligned} &=\frac{-1}{2}\left[\cos 2\left(\frac{\pi}{4}\right)-\cos 2(0)\right] \\\\ &=\frac{-1}{2}\left[\cos \frac{\pi}{2}-\cos 0\right] \end{aligned}

\begin{aligned} &=\frac{-1}{2}[0-1] \\\\ &=\frac{1}{2} \end{aligned}

 

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