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

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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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