Question

${\displaystyle {\int }_0^{\frac{\pi }{4}}\sqrt{1+\sin 2x} \text{d}x}$ ज्ञात कीजिए।

Answer 1

मान लीजिए 

I = ${\displaystyle {\int }_0^{\frac{\pi }{4}}\sqrt{1+\sin 2x}\text{d}x}$

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}\sqrt{{(\sin x+\cos x)}^2}\text{d}x}$

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}(\sin x+\cos x)\text{d}x}$

= ${\displaystyle {(-\cos x+\sin x)}_0^{\frac{\pi }{4}}}$

I = 1.