Solve: _ 0 ^ 2 1+ 2xdx = 1. 1 2 2. 1 3. 2 4. 3 2 - Vidyadip

Question

Solve: $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$

$\frac{1}{2}$
$1$
$2$
$\frac{3}{2}$

✓

Answer

$2$

Solution:
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{xdx}}$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$\int\limits_{0}^{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
$=[-\cos\text{x}+\sin\text{x}]\frac{\frac{\pi}{2}}{{0}}$
$=\Big[-\Big(\cos\frac{\pi}{2}-\cos0\Big)+\Big(\sin\frac{\pi}{2}-\sin0\Big)\Big]$
$=\big[-\big(0-1\big)+\big(1-0\big)\big]$
$=1+1=2$

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