Evaluate integral of sin(x) to the power of two with respect to x

$\int \sin^{2} (x) d x$

Use the half-angle formula to Should be rewritten $\sin^{2} (x)$ as $\frac{1 - \cos (2 x)}{2}$ .

$\int \frac{1 - \cos (2 x)}{2} d x$

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int 1 - \cos (2 x) d x$

Split the single integral into multiple integrals.

$\frac{1}{2} (\int 1 d x + \int - \cos (2 x) d x)$

Since $1$ is constant with respect to $x$ , move $1$ out of the integral.

$\frac{1}{2} (x + C + \int - \cos (2 x) d x)$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$\frac{1}{2} (x + C - \int \cos (2 x) d x)$

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Rewrite.

$\frac{2}{1}$

First, divide $2$ by $1$ .

$2$

Rewrite the problem using $u$ and $d u$ .

$\frac{1}{2} (x + C - \int \cos (u) \frac{1}{2} d u)$

Add $\cos (u)$ and $\frac{1}{2}$ .

$\frac{1}{2} (x + C - \int \frac{\cos (u)}{2} d u)$

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} (x + C - (\frac{1}{2} \int \cos (u) d u))$

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u) + C))$

Simplify.

$\frac{1}{2} (x - \frac{1}{2} \sin (u)) + C$

Replace all occurrences of $u$ with $2 x$ .

$\frac{1}{2} (x - \frac{1}{2} \sin (2 x)) + C$

Clarify the answer.

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Combine $\sin (2 x)$ and $\frac{1}{2}$ .

$\frac{1}{2} (x - \frac{\sin (2 x)}{2}) + C$

Use the distributive law.

$\frac{1}{2} x + \frac{1}{2} (- \frac{\sin (2 x)}{2}) + C$

Combine $\frac{1}{2}$ and $x$ .

$\frac{x}{2} + \frac{1}{2} (- \frac{\sin (2 x)}{2}) + C$

First, multiply $\frac{1}{2} (- \frac{\sin (2 x)}{2})$ .

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Multiply $\frac{1}{2}$ and $\frac{\sin (2 x)}{2}$ .

$\frac{x}{2} - \frac{\sin (2 x)}{2 \cdot 2} + C$

First, multiply $2$ by $2$ .

$\frac{x}{2} - \frac{\sin (2 x)}{4} + C$

Shift terms.

$\frac{1}{2} x - \frac{1}{4} \sin (2 x) + C$

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