What is the integral of ${sin}^{5} (x)$ $sin^5(x)$ ?

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What is the integral of ${sin}^{5} (x)$ $sin^5(x)$ ?

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Answer 1 · 2018-03-30T01:14:40

$\int {sin}^{5} x d x = {sin}^{2} x - sin x - \frac{1}{3} {sin}^{3} x + C$ $intsin^5xdx=sin^2x-sinx-1/3sin^3x+C$

Explanation:

We want $\int {sin}^{5} x d x$ $intsin^5xdx$

Rewrite the integrand as $sin x {({sin}^{2} x)}^{2}$ $sinx(sin^2x)^2$ :

$\int sin x {({sin}^{2} x)}^{2} d x$ $intsinx(sin^2x)^2dx$

Recall the identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ . The identity also tells us that

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$

$\int sin x {(1 - {cos}^{2} x)}^{2} d x$ $intsinx(1-cos^2x)^2dx$

This can be solved using a simple substitution. Let

$u = cos x, d u = - sin x d x, - d u = sin x d x$ $u=cosx, du=-sinxdx, -du=sinxdx$

Rewrite the integral:

$- \int {(1 - u^{2})}^{2} = - \int (1 - 2 u + u^{2})$ $-int(1-u^2)^2=-int(1-2u+u^2)$

Integrate:

$- \int (1 - 2 u + u^{2}) = - (u - u^{2} + \frac{1}{3} u^{3}) + C = u^{2} - u - \frac{1}{3} u^{3} + C$ $-int(1-2u+u^2)=-(u-u^2+1/3u^3)+C=u^2-u-1/3u^3+C$

Rewriting in terms of $x$ $x$ yields

$\int {sin}^{5} x d x = {sin}^{2} x - sin x - \frac{1}{3} {sin}^{3} x + C$ $intsin^5xdx=sin^2x-sinx-1/3sin^3x+C$

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