Calculus Examples

$\int \frac{\sec (x) \tan (x)}{\sec (x) - 1} d x$

Step 1

Let $u = \sec (x) - 1$ . Then $d u = \sec (x) \tan (x) d x$ , so $\frac{1}{\sec (x) \tan (x)} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \sec (x) - 1$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $\sec (x) - 1$ .

$\frac{d}{d x} [\sec (x) - 1]$

Step 1.1.2

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

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

Step 1.1.3

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$\sec (x) \tan (x) + \frac{d}{d x} [- 1]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

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

$\sec (x) \tan (x) + 0$

Step 1.1.4.2

Add $\sec (x) \tan (x)$ and $0$ .

$\sec (x) \tan (x)$

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 $\sec (x) - 1$ .

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