Mathematics

Solve:

$$\displaystyle \int_{0}^{\pi/2}\sqrt{1+sin2x}\ dx =$$ 


ANSWER

$$2$$


SOLUTION
$$\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{2x}}dx}$$

$$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{{\sin}^{2}{x}+{\cos}^{2}{x}+2\sin{x}\cos{x}}dx}$$

$$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{{\left(\sin{x}+\cos{x}\right)}^{2}}dx}$$

$$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(\sin{x}+\cos{x}\right)dx}$$

$$=\left[-\cos{x}+\sin{x}\right]_{0}^{\frac{\pi}{2}}$$

$$=\left[-\left(\cos{\dfrac{\pi}{2}}-\cos{0}\right)+\left(\sin{\dfrac{\pi}{2}}-\sin{0}\right)\right]$$

$$=\left[-\left(0-1\right)+\left(1-0\right)\right]$$

$$=1+1=2$$
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Single Correct Medium Published on 17th 09, 2020
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