How do you integrate $\int {sin (x)}^{3} \cdot cos (x) d x$ $int sin(x)^3 * cos(x) dx$ ?

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How do you integrate $\int {sin (x)}^{3} \cdot cos (x) d x$ $int sin(x)^3 * cos(x) dx$ ?

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Answer 1 · 2016-05-01T21:09:25

Use a $u$ $u$ -substitution to get $\int {sin}^{3} x cos x d x = \frac{{sin}^{4} x}{4} + C$ $intsin^3xcosxdx=sin^4x/4+C$ .

Explanation:

What we have in this integral is a function, $sin x$ $sinx$ , and its derivative, $cos x$ $cosx$ . That means the integral is solvable using a $u$ $u$ -substitution:
Let $u = sin x \to \frac{d u}{d x} = cos x \to d u = cos x d x$ $u=sinx->(du)/dx=cosx->du=cosxdx$
With this substitution, $\int {sin}^{3} x cos x d x$ $intsin^3xcosxdx$ becomes:
$\int u^{3} d u$ $intu^3du$

This new integral is easily evaluated using the reverse power rule:
$\int u^{3} d u = \frac{u^{3 + 1}}{3 + 1} + C = \frac{u^{4}}{4} + C$ $intu^3du=(u^(3+1))/(3+1)+C=u^4/4+C$

Because $u = sin x$ $u=sinx$ , we can substitute to get a final answer of:
$\int {sin}^{3} x cos x d x = \frac{{sin}^{4} x}{4} + C$ $intsin^3xcosxdx=sin^4x/4+C$

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How do you integrate $\int {sin (x)}^{3} \cdot cos (x) d x$ $int sin(x)^3 * cos(x) dx$ ?

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Explanation:

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How do you integrate $\int {sin (x)}^{3} \cdot cos (x) d x$ $int sin(x)^3 * cos(x) dx$ ?