# Integration of Sin Squared x

In this tutorial we shall derive the integral of sine squared x.

The integration is of the form

\[I = \int {{{\sin }^2}xdx} \]

This integral cannot be evaluated by the direct formula of integration, so using the trigonometric identity of half angle $${\sin ^2}x = \frac{{1 – \cos 2x}}{2}$$, we have

\[\begin{gathered} I = \int {\left( {\frac{{1 – \cos 2x}}{2}} \right)dx} \\ \Rightarrow I = \frac{1}{2}\int {\left( {1 – \cos 2x} \right)dx} \\ \Rightarrow I = \frac{1}{2}\int {1dx – \frac{1}{2}\int {\cos 2xdx} } \\ \end{gathered} \]

Using the integral formula $$\int {\cos kxdx = \frac{{\sin kx}}{k} + c} $$, we have

\[\begin{gathered} \int {{{\sin }^2}x} dx = \frac{1}{2}x – \frac{1}{2}\frac{{\sin 2x}}{2} + c \\ \Rightarrow \int {{{\sin }^2}x} dx = \frac{1}{2}x – \frac{1}{4}\sin 2x + c \\ \end{gathered} \]

JAMES DALTON

July 21@ 5:11 pmEvaluating from zero to 2pi gives pi but I thought the answer should be one half. Where I am I going wrong?

Mark Huth

December 28@ 5:01 amThe correct answer is pi. If you thought it should be 1/2, then that is where you went wrong. Graph the double hump of the sine^2 and you can easily see that 1/2 is not correct.