How do you find the antiderivative of $\int sin x {(cos x)}^{\frac{3}{2}} d x$ $int sinx(cosx)^(3/2) dx$ ?

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How do you find the antiderivative of $\int sin x {(cos x)}^{\frac{3}{2}} d x$ $int sinx(cosx)^(3/2) dx$ ?

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Answer 1 · 2017-01-14T00:54:34

$- \frac{2}{5} {(cos x)}^{\frac{5}{2}} + C$ $-2/5(cosx)^(5/2)+C$

Explanation:

Since we don't want to have to deal with the $3 / 2$ $3//2$ power, let $u = cos x$ $u=cosx$ . This implies that $d u = - sin x . d x$ $du=-sinxcolor(white).dx$ .

So, we need to modify our integral just a little:

$\int sin x {(cos x)}^{\frac{3}{2}} d x = - \int {(cos x)}^{\frac{3}{2}} (- sin x . d x) = - \int u^{\frac{3}{2}} . d u$ $intsinx(cosx)^(3/2)dx=-int(cosx)^(3/2)(-sinxcolor(white).dx)=-intu^(3/2)color(white).du$

Integrate this using the rule $\int u^{n} . d u = \frac{u^{n + 1}}{n + 1}$ $intu^ncolor(white).du=u^(n+1)/(n+1)$ , where $n \neq - 1$ $n!=-1$ .

$= - \frac{u^{\frac{5}{2}}}{\frac{5}{2}} = - \frac{2}{5} u^{\frac{5}{2}} = - \frac{2}{5} {(cos x)}^{\frac{5}{2}} + C$ $=-u^(5/2)/(5/2)=-2/5u^(5/2)=-2/5(cosx)^(5/2)+C$

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