Calculus Examples

$\int \sin (9 t) + \sin (2 t) d t$

Step 1

Split the single integral into multiple integrals.

$\int \sin (9 t) d t + \int \sin (2 t) d t$

Step 2

Let $u_{1} = 9 t$ . Then $d u_{1} = 9 d t$ , so $\frac{1}{9} d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.1

Let $u_{1} = 9 t$ . Find $\frac{d u_{1}}{d t}$ .

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Step 2.1.1

Differentiate $9 t$ .

$\frac{d}{d t} [9 t]$

Step 2.1.2

Since $9$ is constant with respect to $t$ , the derivative of $9 t$ with respect to $t$ is $9 \frac{d}{d t} [t]$ .

$9 \frac{d}{d t} [t]$

Step 2.1.3

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

$9 \cdot 1$

Step 2.1.4

Multiply $9$ by $1$ .

$9$

Step 2.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \sin (u_{1}) \frac{1}{9} d u_{1} + \int \sin (2 t) d t$

Step 3

Combine $\sin (u_{1})$ and $\frac{1}{9}$ .

$\int \frac{\sin (u_{1})}{9} d u_{1} + \int \sin (2 t) d t$

Step 4

Since $\frac{1}{9}$ is constant with respect to $u_{1}$ , move $\frac{1}{9}$ out of the integral.

$\frac{1}{9} \int \sin (u_{1}) d u_{1} + \int \sin (2 t) d t$

Step 5

The integral of $\sin (u_{1})$ with respect to $u_{1}$ is $- \cos (u_{1})$ .

$\frac{1}{9} (- \cos (u_{1}) + C) + \int \sin (2 t) d t$

Step 6

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

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Step 6.1

Let $u_{2} = 2 t$ . Find $\frac{d u_{2}}{d t}$ .

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Step 6.1.1

Differentiate $2 t$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

$2 \cdot 1$

Step 6.1.4

Multiply $2$ by $1$ .

$2$

Step 6.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{9} (- \cos (u_{1}) + C) + \int \sin (u_{2}) \frac{1}{2} d u_{2}$

Step 7

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

$\frac{1}{9} (- \cos (u_{1}) + C) + \int \frac{\sin (u_{2})}{2} d u_{2}$

Step 8

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

$\frac{1}{9} (- \cos (u_{1}) + C) + \frac{1}{2} \int \sin (u_{2}) d u_{2}$

Step 9

The integral of $\sin (u_{2})$ with respect to $u_{2}$ is $- \cos (u_{2})$ .

$\frac{1}{9} (- \cos (u_{1}) + C) + \frac{1}{2} (- \cos (u_{2}) + C)$

Step 10

Simplify.

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Step 10.1

Simplify.

$- \frac{\cos (u_{1})}{9} + \frac{1}{2} (- \cos (u_{2})) + C$

Step 10.2

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

$- \frac{\cos (u_{1})}{9} - \frac{\cos (u_{2})}{2} + C$

Step 11

Substitute back in for each integration substitution variable.

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Step 11.1

Replace all occurrences of $u_{1}$ with $9 t$ .

$- \frac{\cos (9 t)}{9} - \frac{\cos (u_{2})}{2} + C$

Step 11.2

Replace all occurrences of $u_{2}$ with $2 t$ .

$- \frac{\cos (9 t)}{9} - \frac{\cos (2 t)}{2} + C$

Step 12

Reorder terms.

$- \frac{1}{9} \cos (9 t) - \frac{1}{2} \cos (2 t) + C$