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 $\frac{1}{\sin (x)} d u = d x$ . 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

Evaluate $\frac{d}{d x} [- \cos (x)]$ .

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Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

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

Step 1.1.3.2

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

$0 - - \sin (x)$

Step 1.1.3.3

Multiply $- 1$ by $- 1$ .

$0 + 1 \sin (x)$

Step 1.1.3.4

Multiply $\sin (x)$ by $1$ .

$0 + \sin (x)$

Step 1.1.4

Add $0$ and $\sin (x)$ .

$\sin (x)$

Step 1.2

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

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

Step 2

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

$\ln (| u |) + C$

Step 3

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

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