Calculus Examples

$\int_{0}^{π} \cos (2 x) d x$

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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Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1$

Multiply $2$ 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 $u = 2 x$ .

$u_{upper} = 2 π$

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

$u_{lower} = 0 u_{upper} = 2 π$

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

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

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

$\int_{0}^{2 π} \frac{\cos (u)}{2} d u$

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

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

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

$\frac{1}{2} \sin (u)]_{0}^{2 π}$

Evaluate $\sin (u)$ at $2 π$ and at $0$ .

$\frac{1}{2} (\sin (2 π) - \sin (0))$

Simplify.

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

$\frac{1}{2} (\sin (2 π) - 0)$

Multiply $- 1$ by $0$ .

$\frac{1}{2} (\sin (2 π) + 0)$

Add $\sin (2 π)$ and $0$ .

$\frac{1}{2} \sin (2 π)$

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

$\frac{\sin (2 π)}{2}$

Simplify.

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Simplify the numerator.

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Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{\sin (0)}{2}$

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

$\frac{0}{2}$

Divide $0$ by $2$ .

$0$