Question

Evaluate :

${\int }_0^{\pi /4}\sqrt{1+\sin 2x}dx$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$I={\int }_0^{\pi /4}\sqrt{1+\sin 2x}dx$

$I={\int }_0^{\pi /4}\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx$

$I={\int }_0^{\pi /4}\sqrt{(\sin x+\cos x{)}^2}dx$

$I={\int }_0^{\pi /4}|\cos x+\sin x|dx$

$I={\int }_0^{\pi /4}(\cos x+\sin x)dx$

$I=[\sin x-\cos x{]}_0^{\pi /4}$

$I=(\sin \frac{\pi }{4}-\cos \frac{\pi }{4})-(\sin 0-\cos 0)$

$I=(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})-(0-1)=1$