What is the antiderivative of $(sin x) \cdot (cos x)$ $(sinx)*(cosx)$ ?

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Answer 1 · 2018-02-05T14:33:41

$\int sin x cos x d x = - \frac{1}{2} cos 2 x$ $intsinxcosxdx=-1/2cos2x$

Antiderivative of $sin x cos x$ $sinxcosx$ means

$\int sin x cos x d x$ $intsinxcosxdx$

= $\int sin 2 x d x$ $intsin2xdx$ and let $u = 2 x$ $u=2x$ then $d u = \frac{d x}{2}$ $du=(dx)/2$

and hence $\int sin 2 x d x = \frac{1}{2} \int sin u d u$ $intsin2xdx=1/2intsinudu$

= $- \frac{1}{2} cos u = - \frac{1}{2} cos 2 x$ $-1/2cosu=-1/2cos2x$

What is the antiderivative of (sinx)*(cosx)(sinx)⋅(cosx)(sinx)*(cosx)?