Calculus Examples

$\int \frac{\sin (x)}{1 + \cos (x)} d x$

Step 1

Let $u = 1 + \cos (x)$ . 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 1.1

Let $u = 1 + \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $1 + \cos (x)$ .

$\frac{d}{d x} [1 + \cos (x)]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\cos (x)]$

Step 1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 1.1.3

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

$0 - \sin (x)$

Step 1.1.4

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

$- \sin (x)$

Step 1.2

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

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

Step 2

Split the fraction into multiple fractions.

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

Step 3

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

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

Step 4

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

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

Step 5

Simplify.

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

Step 6

Replace all occurrences of $u$ with $1 + \cos (x)$ .

$- \ln (| 1 + \cos (x) |) + C$