How do you integrate $\int {cos}^{3} x sin x d x$ $int cos^3xsinxdx$ ?

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How do you integrate $\int {cos}^{3} x sin x d x$ $int cos^3xsinxdx$ ?

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Answer 1 · 2017-01-23T00:06:30

$\int {cos}^{3} x sin x d x = - \frac{{cos}^{4} x}{4} + C$ $int cos^3x sinx dx = - cos^4x/4+C$

Explanation:

Note that:

$- sin x d x = d (cos x)$ $-sinx dx = d(cosx)$

so if you substitute $u = cos x$ $u= cosx$ you have:

$\int {cos}^{3} x sin x d x = - \int u^{3} d u = - \frac{u^{4}}{4} + C = - \frac{{cos}^{4} x}{4} + C$ $int cos^3x sinx dx = - int u^3 du = - u^4/4 +C = - cos^4x/4+C$

Answer 2 · 2017-01-23T01:49:21

Since $\frac{d}{d x} (cos (x)) = - sin (x)$ $d/dx(cos(x))=-sin(x)$

then

$\int {cos}^{3} (x) sin (x) d x = - \int {cos}^{3} (x) d cos (x) = - \frac{{cos}^{4} (x)}{4} + C$ $intcos^3(x)sin(x)dx=-intcos^3(x)dcos(x)=-(cos^(4)(x))/4+C$

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