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\int \sin ^{2}{x}\,dx\,

To compute this integral, we must first use trig identities:

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

\sin ^{2}{x}-\cos ^{2}{x}=-\cos 2x\,

Add the second equation to the first.

2\sin ^{2}{x}=1-\cos 2x\,


\sin ^{2}{x}={\frac  {1}{2}}-{\frac  {\cos 2x}{2}}

Substitute this identity into the integral

\int {{\frac  {1}{2}}-{\frac  {\cos 2x}{2}}dx}

This can be split into two pieces and evaluated separately

\int {{\frac  {1}{2}}dx}-\int {{\frac  {\cos 2x}{2}}dx}

The first is trivial

\int {{\frac  {1}{2}}dx}={\frac  {x}{2}}+C_{1} The second can be computed by simple substitution 2x=u;dx={\frac  {du}{2}}\,

\int {{\frac  {\cos u}{4}}du}={\frac  {\sin u}{4}}+C_{2}

Finally, we complete our substitution and combine our solutions

{\frac  {x}{2}}-{\frac  {\sin 2x}{4}}+C, where C=C_{1}-C_{2}\,

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