Evaluate integral from 0 to 2pi of cos(x)^2 with respect to x

$\int_{0}^{2 π} \cos^{2} (x) d x$

Use the half–angle formula to rewrite $\cos^{2} (x)$ as $\frac{1 + \cos (2 x)}{2}$ .

$\int_{0}^{2 π} \frac{1 + \cos (2 x)}{2} d x$

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

$\frac{1}{2} \int_{0}^{2 π} 1 + \cos (2 x) d x$

Split the single integral into multiple integrals.

$\frac{1}{2} (\int_{0}^{2 π} 1 d x + \int_{0}^{2 π} \cos (2 x) d x)$

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

$\frac{1}{2} (x]_{0}^{2 π} + \int_{0}^{2 π} \cos (2 x) d x)$

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.

$\frac{2}{1}$

Divide $2$ by $1$ .

$2$

Substitute the lower limit in for $x$ in $u = 2 x$ .

$u_{lower} = 2 \cdot 0$

Multiply $2$ by $0$ .

$u_{lower} = 0$

Substitute the upper limit in for $x$ in $u = 2 x$ .

$u_{upper} = 2 (2 π)$

Multiply $2$ by $2$ .

$u_{upper} = 4 π$

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0 u_{upper} = 4 π$

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\frac{1}{2} (x]_{0}^{2 π} + \int_{0}^{4 π} \cos (u) \frac{1}{2} d u)$

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

$\frac{1}{2} (x]_{0}^{2 π} + \int_{0}^{4 π} \frac{\cos (u)}{2} d u)$

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

$\frac{1}{2} (x]_{0}^{2 π} + \frac{1}{2} \int_{0}^{4 π} \cos (u) d u)$

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

$\frac{1}{2} (x]_{0}^{2 π} + \frac{1}{2} \sin (u)]_{0}^{4 π})$

Combine $\frac{1}{2}$ and $\sin (u)]_{0}^{4 π}$ .

$\frac{1}{2} (x]_{0}^{2 π} + \frac{\sin (u)]_{0}^{4 π}}{2})$

Substitute and simplify.

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Evaluate $x$ at $2 π$ and at $0$ .

$\frac{1}{2} ((2 π) + 0 + \frac{\sin (u)]_{0}^{4 π}}{2})$

Evaluate $\sin (u)$ at $4 π$ and at $0$ .

$\frac{1}{2} ((2 π) + 0 + \frac{(\sin (4 π)) - \sin (0)}{2})$

Add $2 π$ and $0$ .

$\frac{1}{2} (2 π + \frac{\sin (4 π) - \sin (0)}{2})$

Simplify.

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The exact value of $\sin (0)$ is $0$ .

$\frac{1}{2} (2 π + \frac{\sin (4 π) - 0}{2})$

Multiply $- 1$ by $0$ .

$\frac{1}{2} (2 π + \frac{\sin (4 π) + 0}{2})$

Add $\sin (4 π)$ and $0$ .

$\frac{1}{2} (2 π + \frac{\sin (4 π)}{2})$

Simplify the answer.

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Simplify each term.

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Simplify the numerator.

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$4 π$ is a full rotation so replace with $0$ .

$\frac{1}{2} (2 π + \frac{\sin (0)}{2})$

The exact value of $\sin (0)$ is $0$ .

$\frac{1}{2} (2 π + \frac{0}{2})$

Divide $0$ by $2$ .

$\frac{1}{2} (2 π + 0)$

Add $2 π$ and $0$ .

$\frac{1}{2} (2 π)$

Cancel the common factor of $2$ .

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Factor $2$ out of $2 π$ .

$\frac{1}{2} (2 (π))$

Cancel the common factor.

$\frac{1}{2} (2 π)$

Rewrite the expression.

$π$

The result can be shown in multiple forms.

Exact Form:

$π$

Decimal Form:

$3.14159265 \dots$

Evaluate integral from 0 to 2pi of cos(x)^2 with respect to x

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