# Find the antiderivative of (\sin x)^{2}

Find the antiderivative of ${\left(\mathrm{sin}x\right)}^{2}$
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eskalopit
$\int {\mathrm{sin}}^{2}xdx$ has no clear solutions.
Instead of integrating ${\mathrm{sin}}^{2}x$, you can integrate $\frac{1}{2}\left(1-\mathrm{cos}2x\right)$
$\int \frac{1}{2}\left(1-\mathrm{cos}2x\right)dx$
$=\frac{1}{2}\int 1-\mathrm{cos}2xdx$
Now, use the sum rule to split this into:
$\frac{1}{2}\int 1dx-\frac{1}{2}\int \mathrm{cos}2xdx$
$\frac{1}{2}x-\frac{1}{2}\int \mathrm{cos}2xdx$
$\int \mathrm{cos}2xdx=\frac{1}{2}\mathrm{sin}2x$
So, we have $\frac{1}{2}x-\frac{1}{2}\int \mathrm{cos}2xdx$
$\frac{1}{2}x-\frac{1}{2}\left(\frac{1}{2}\mathrm{sin}2x\right)+C=\frac{1}{2}x-\frac{1}{4}\mathrm{sin}2x+C$
###### Not exactly what you’re looking for?
Wendy Boykin
$\int {\left(\mathrm{sin}\left(x\right)\right)}^{2}dx$
$\int {\mathrm{sin}}^{2}\left(x\right)dx$
Rewrite using trig identities
$\int \frac{1-\mathrm{cos}\left(2x\right)}{2}dx$
Take the constant out
$=\frac{1}{2}×\int 1-\mathrm{cos}\left(2x\right)dx$
Apply the sum rule
$=\frac{1}{2}\left(\int 1dx-\int \mathrm{cos}\left(2x\right)dx\right)$
As we know that $\int 1dx=x$, $\int \mathrm{cos}\left(2x\right)dx=\frac{1}{2}\mathrm{sin}\left(2x\right)$, we have
$=\frac{1}{2}\left(x-\frac{1}{2}\mathrm{sin}\left(2x\right)\right)$
$=\frac{1}{2}\left(x-\frac{1}{2}\mathrm{sin}\left(2x\right)\right)+C$