Question

$\int (\sin x-\cos x{)}^{2}dx$.

Answer 1

We are given that,

$I=\int (sinxx-cosx{)}^{2}dx$

$=\int (si{n}^{2}x-2sinx.cosx+co{s}^{2}x)dx$

(Expand it )

$=\int (si{n}^{2}x+co{s}^{2}x-2sinxco{s}^{2}x)dx$

as we know, $si{n}^{2}x+co{s}^{2}x=1\&2sinxcosx=sin2x$

$I=\int (1-sin2x)dx$

$=\int dx-\int sin2xdx$

$=x-\left(\frac{-cos2x}{2}\right)+c$

$I=x+\frac{cosx}{2}+c$

where c is any arbitary constant.