How do you integrate $\int {(sin x + cos x)}^{2}$ $int(sinx+cosx)^2$ ?

Question

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Answer 1 · 2015-03-01T16:15:34

The answer is: $x - \frac{1}{2} cos 2 x + c$ $x-1/2cos2x+c$ .

$\int {(sin x + cos x)}^{2} d x = \int ({sin}^{2} x + 2 sin x cos x + {cos}^{2} x) d x =$ $int(sinx+cosx)^2dx=int(sin^2x+2sinxcosx+cos^2x)dx=$

$= \int (1 + sin 2 x) d x = \int d x + \frac{1}{2} \int 2 sin 2 x d x =$ $=int(1+sin2x)dx=intdx+1/2int2sin2xdx=$

$= x - \frac{1}{2} cos 2 x + c$ $=x-1/2cos2x+c$ .

How do you integrate int(sinx+cosx)^2∫(sinx+cosx)2int(sinx+cosx)^2?