Question

Friends Solve This
$\huge{ \red {\tt{\int \frac{((sin x + cos x) ^ 2)}{(sqrt(1 + sin 2x))} dx \: from \: 0 \: to \: \pi / 2 =}}}$
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Answer 1

Step-by-step explanation:

${\int_0 }^{ \frac{\pi}{2} } \frac{ {( \sin x + \cos x ) }^{2} }{ \sqrt{1 + \sin 2x} } dx \\= {\int_0 }^{ \frac{\pi}{2} } \frac{ {( \sin x + \cos x ) }^{2} }{ \sqrt{ {\sin }^{2}x + { \cos}^{2} x + 2 \sin x. \cos x } } dx \\= {\int_0 }^{ \frac{\pi}{2} } \frac{ {( \sin x + \cos x ) }^{2} }{ \sqrt{{(\sin x + \cos x ) }^{2}} } dx \\= {\int_0 }^{ \frac{\pi}{2} } \frac{ {( \sin x + \cos x ) }^{2} }{ (\sin x + \cos x ) } dx \\= {\int_0 }^{ \frac{\pi}{2} } ( \sin x + \cos x ) dx \\= {\int_0 }^{ \frac{\pi}{2} } \sin xdx + {\int_0 }^{ \frac{\pi}{2} } \cos xdx\\ = {[ - \cos x ]}^{ \frac{\pi}{2} }_0 + {[ \sin x ]}^{ \frac{\pi}{2} }_0 \\ = [ - \cos \frac{\pi}{2} - ( - \cos 0)] + [ \sin \frac{\pi}{2} - \sin 0] \\ = [ - 0 + 1] + [ 1 - 0] \\ = 2 \\ \\\red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

no!

how can a community with 59% be in minority

pls try again