Question

Evaluate $\int \sqrt{1+\sin 2x}dx,$ where $x\in (\frac{\pi }{2},\frac{3\pi }{4})$

• A. $\sin x-cosx+c$
• B. $\sin x+\cos x+c$
• C. $\cos x-\sin x+c$
• D. $\frac{(sinx+cosx{)}^2}{2}+c$

Answer 1

Answer: (A)

$\sin x-cosx+c$

The expression $1+\sin 2x$ can be simplified and written as,

$1+\sin 2x={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x$

= ${(\sin x+\cos x)}^2$

Now $\int \sqrt{(1+\sin 2x)}$ = $\int \sqrt{{(\sin x+\cos x)}^2}$

$|\sin x+\cos x|=\sqrt{2}\left|\sin (x+\frac{\pi }{4})\right|$

$\sin (x+\frac{\pi }{4})\ge 0,\forall x\in [-\frac{\pi }{4},\frac{3\pi }{4}]$

Hence, in the given interval, $\sqrt{2}\left|\sin (x+\frac{\pi }{4})\right|=\sqrt{2}\sin (x+\frac{\pi }{4})=\cos x+\sin x$

Hence, $\int (\cos x+\sin x)dx=\sin x-\cos x+c$