Question

Evaluate the integral
${\int }_0^{\pi /2}\sqrt{1+\sin x}dx$

• A. $1$
• B. $4$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

${\int }_0^{\frac{\pi }{2}}\sqrt{1+sinx}dx$

$\int \sqrt{1+sinx}dx=\int (cos\frac{x}{2}+sin\frac{x}{2})dx$

$=2+(sin\frac{x}{2}-cos\frac{x}{2})+c$

$=2(sin\frac{x}{2}-cos\frac{x}{2})+c$

${\int }_0^{\frac{\pi }{2}}\sqrt{1+sinx}dx={[2+(sin\frac{x}{2}-cos\frac{x}{2})+c]}_0^{\frac{\pi }{2}}$

$=\left[2\right(sin\frac{\pi }{4}-cos\frac{\pi }{4})+c]-\left[2\right(sin0-cos0)+c]$

$=2\left(0\right)-2(-1)=2$

${\int }_0^{\frac{\pi }{2}}\sqrt{1+sinx}dx=2$