Mathematics

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

$2$

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
Questions 203525
Subjects 9
Chapters 126
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