Calculus Examples

$\int \frac{4 \sin (x)}{3 + \cos (x)} d x$

Step 1

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$4 \int \frac{\sin (x)}{3 + \cos (x)} d x$

Step 2

Let $u = 3 + \cos (x)$ . Then $d u = - \sin (x) d x$ , so $- d u = \sin (x) d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = 3 + \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $3 + \cos (x)$ .

$\frac{d}{d x} [3 + \cos (x)]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $3 + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [\cos (x)]$ .

$\frac{d}{d x} [3] + \frac{d}{d x} [\cos (x)]$

Step 2.1.2.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\cos (x)]$

Step 2.1.3

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$0 - \sin (x)$

Step 2.1.4

Subtract $\sin (x)$ from $0$ .

$- \sin (x)$

Step 2.2

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

$4 \int \frac{- 1}{u} d u$

Step 3

Move the negative in front of the fraction.

$4 \int - \frac{1}{u} d u$

Step 4

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

$4 (- \int \frac{1}{u} d u)$

Step 5

Multiply $- 1$ by $4$ .

$- 4 \int \frac{1}{u} d u$

Step 6

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$- 4 (\ln (| u |) + C)$

Step 7

Simplify.

$- 4 \ln (| u |) + C$

Step 8

Replace all occurrences of $u$ with $3 + \cos (x)$ .

$- 4 \ln (| 3 + \cos (x) |) + C$