# What is the integral of int sin^5 (x) dx?

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#### Explanation

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#### Explanation:

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20
Mar 4, 2018

The answer is $= - \frac{1}{5} {\cos}^{5} x + \frac{2}{3} {\cos}^{3} x - \cos x + C$

#### Explanation:

We need

${\sin}^{2} x + {\cos}^{2} x = 1$

The integral is

$\int {\sin}^{5} \mathrm{dx} = \int {\left(1 - {\cos}^{2} x\right)}^{2} \sin x \mathrm{dx}$

Perform the substitution

$u = \cos x$, $\implies$, $\mathrm{du} = - \sin x \mathrm{dx}$

Therefore,

$\int {\sin}^{5} \mathrm{dx} = - \int {\left(1 - {u}^{2}\right)}^{2} \mathrm{du}$

$= - \int \left(1 - 2 {u}^{2} + {u}^{4}\right) \mathrm{du}$

$= - \int {u}^{4} \mathrm{du} + 2 \int {u}^{2} \mathrm{du} - \int \mathrm{du}$

$= - {u}^{5} / 5 + 2 {u}^{3} / 3 - u$

$= - \frac{1}{5} {\cos}^{5} x + \frac{2}{3} {\cos}^{3} x - \cos x + C$

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