Calculus Examples

$\frac{d y}{d x} = \sin (9 x)$

Step 1

Rewrite the equation.

$d y = \sin (9 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \sin (9 x) d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \sin (9 x) d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

Let $u = 9 x$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $9 x$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

$9 \cdot 1$

Step 2.3.1.1.4

Multiply $9$ by $1$ .

$9$

Step 2.3.1.2

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

$y + C_{1} = \int \sin (u) \frac{1}{9} d u$

Step 2.3.2

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

$y + C_{1} = \int \frac{\sin (u)}{9} d u$

Step 2.3.3

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

$y + C_{1} = \frac{1}{9} \int \sin (u) d u$

Step 2.3.4

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

$y + C_{1} = \frac{1}{9} (- \cos (u) + C_{2})$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Simplify.

$y + C_{1} = \frac{1}{9} (- \cos (u)) + C_{2}$

Step 2.3.5.2

Combine $\frac{1}{9}$ and $\cos (u)$ .

$y + C_{1} = - \frac{\cos (u)}{9} + C_{2}$

Step 2.3.6

Replace all occurrences of $u$ with $9 x$ .

$y + C_{1} = - \frac{\cos (9 x)}{9} + C_{2}$

Step 2.3.7

Reorder terms.

$y + C_{1} = - \frac{1}{9} \cos (9 x) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$y = - \frac{1}{9} \cos (9 x) + K$