How to find the indefinite integral of sin^2x

In this post, we are going to find the indefinite integral of sin^2xdx.

Given the integral,

$\displaystyle \int \sin^2xdx$

Recall the double angle identity,

$\cos(2x)=\cos^2x-\sin^2x\\ \textup{but }\sin^2x+\cos^2x=1 \to \cos^2x=1-\sin^2x$

This simplifies the identity to,

$\begin {align*} \cos(2x)&=1-\sin^2x-\sin^2x\\ \cos(2x)&=1-2\sin^2x\\ 2\sin^2x&=1-\cos(2x)\\ \sin^2x&=\dfrac{1}{2}-\dfrac{\cos(2x)}{2} \end {align*}$

Going back to the original integral,

$\begin {align*} \displaystyle \int \sin^2xdx&=\displaystyle \int \left (\dfrac{1}{2}-\dfrac{\cos(2x)}{2} \right )dx\\ \displaystyle \int \sin^2xdx&=\displaystyle \int \left (\dfrac{1}{2}\right )dx-\left (\dfrac{\cos(2x)}{2} \right )dx\\ \displaystyle \int \sin^2xdx&=\dfrac{1}{2}\displaystyle \int dx-\dfrac{1}{2}\displaystyle \int \cos(2x) dx\\ \end {align*}$

Let u=2x, du=2dx,

$\begin {align*} \displaystyle \int \sin^2xdx&=\dfrac{1}{2}\displaystyle \int dx-\dfrac{1}{2}\cdot \dfrac{1}{2}\displaystyle \int \cos(2x) 2dx\\ \displaystyle \int \sin^2xdx&=\dfrac{1}{2}\displaystyle \int dx-\dfrac{1}{2}\cdot \dfrac{1}{2}\displaystyle \int \cos u du\\ \displaystyle \int \sin^2xdx&=\dfrac{1}{2}x-\dfrac{1}{4}(\sin u)\\ \displaystyle \int \sin^2xdx&=\dfrac{1}{2}x-\dfrac{1}{4}\sin 2x+c\\ \end {align*}$

How to find the indefinite integral of sin^2x

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How to find the indefinite integral of sin^2x

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