Evaluate \int \sin^{5}xdx.

Evaluate $\int {\mathrm{sin}}^{5}xdx$.
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SchulzD
Step 1
In some cases integrals can be simplified by use of a substitution. This usually results in simplifying the integral in terms of known standard integrals.
For the given integral the integrand can be written as $\mathrm{sin}\left(x\right){\mathrm{sin}}^{4}\left(x\right)=\mathrm{sin}\left(x\right){\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{2}$.
Then the substitution $u=\mathrm{cos}\left(x\right)$ will simplify the integrand expression.
Step 2
The given indefinite integral to be evaluated is $\int {\mathrm{sin}}^{5}\left(x\right)dx$. This can be rewritten as $\int \mathrm{sin}\left(x\right){\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{2}dx$.Use the substitution $u=\mathrm{cos}\left(x\right)$. Differentiating this gives $-du=\mathrm{sin}\left(x\right)dx$.Substitute this in the integral and evaluate the integral.
$\int {\mathrm{sin}}^{5}\left(x\right)dx=\int \mathrm{sin}\left(x\right){\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{2}dx$
$=-\int {\left(1-{u}^{2}\right)}^{2}du$
$=-\int \left(1+{u}^{4}-2{u}^{2}\right)du$
$=-u-\frac{{u}^{5}}{5}+\frac{2{u}^{3}}{3}+C$
$=-\mathrm{cos}\left(x\right)-\frac{{\mathrm{cos}}^{5}\left(x\right)}{5}+\frac{2{\mathrm{cos}}^{3}\left(x\right)}{3}+C$