Question

Integral 0 to pi bye 2 sqrt of (1+sin2x) dx

Answer 1

$I =\int\limits^\frac{\pi }{2} _0 {\sqrt{1+sin2x} } \, dx$

$I =\int\limits^\frac{\pi }{2} _0 {\sqrt{Sin^2x+Cos^2x+2SinxCosx} } \, dx$

$I=\int\limits^\frac{\pi }{2} _0 {\sqrt{(Sinx+Cosx)^2} } \, dx$

$I =\int\limits^\frac{\pi }{2} _0 {(Sinx+Cosx) } \, dx$

$I =\int\limits^\frac{\pi }{2} _0 {Sinx} } \, dx+\int\limits^\frac{\pi }{2} _0 {Cosx} } \, dx$

$I = [-Cosx]\limits^\frac{\pi }{2}_0 +[Sinx]\limits^\frac{\pi }{2}_0$

$I=Cos0-Cos\frac{\pi}{2}+Sin\frac{\pi}{2}-Sin0$

$I = 2$

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