Calculus Examples

$\int \frac{\sec^{2} (8 x)}{\tan (8 x) - 6} d x$

Step 1

Let $u_{2} = \tan (8 x) - 6$ . Then $d u_{2} = 8 \sec^{2} (8 x) d x$ , so $\frac{1}{8} d u_{2} = \sec^{2} (8 x) d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 1.1

Let $u_{2} = \tan (8 x) - 6$ . Find $\frac{d u_{2}}{d x}$ .

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Step 1.1.1

Differentiate $\tan (8 x) - 6$ .

$\frac{d}{d x} [\tan (8 x) - 6]$

Step 1.1.2

By the Sum Rule, the derivative of $\tan (8 x) - 6$ with respect to $x$ is $\frac{d}{d x} [\tan (8 x)] + \frac{d}{d x} [- 6]$ .

$\frac{d}{d x} [\tan (8 x)] + \frac{d}{d x} [- 6]$

Step 1.1.3

Evaluate $\frac{d}{d x} [\tan (8 x)]$ .

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Step 1.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 8 x$ .

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Step 1.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $8 x$ .

$\frac{d}{d u_{1}} [\tan (u_{1})] \frac{d}{d x} [8 x] + \frac{d}{d x} [- 6]$

Step 1.1.3.1.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

$\sec^{2} (u_{1}) \frac{d}{d x} [8 x] + \frac{d}{d x} [- 6]$

Step 1.1.3.1.3

Replace all occurrences of $u_{1}$ with $8 x$ .

$\sec^{2} (8 x) \frac{d}{d x} [8 x] + \frac{d}{d x} [- 6]$

Step 1.1.3.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$\sec^{2} (8 x) (8 \frac{d}{d x} [x]) + \frac{d}{d x} [- 6]$

Step 1.1.3.3

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

$\sec^{2} (8 x) (8 \cdot 1) + \frac{d}{d x} [- 6]$

Step 1.1.3.4

Multiply $8$ by $1$ .

$\sec^{2} (8 x) \cdot 8 + \frac{d}{d x} [- 6]$

Step 1.1.3.5

Move $8$ to the left of $\sec^{2} (8 x)$ .

$8 \sec^{2} (8 x) + \frac{d}{d x} [- 6]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$8 \sec^{2} (8 x) + 0$

Step 1.1.4.2

Add $8 \sec^{2} (8 x)$ and $0$ .

$8 \sec^{2} (8 x)$

Step 1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{u_{2}} \cdot \frac{1}{8} d u_{2}$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{8}$ .

$\int \frac{1}{u_{2} \cdot 8} d u_{2}$

Step 2.2

Move $8$ to the left of $u_{2}$ .

$\int \frac{1}{8 u_{2}} d u_{2}$

Step 3

Since $\frac{1}{8}$ is constant with respect to $u_{2}$ , move $\frac{1}{8}$ out of the integral.

$\frac{1}{8} \int \frac{1}{u_{2}} d u_{2}$

Step 4

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{8} (\ln (| u_{2} |) + C)$

Step 5

Simplify.

$\frac{1}{8} \ln (| u_{2} |) + C$

Step 6

Replace all occurrences of $u_{2}$ with $\tan (8 x) - 6$ .

$\frac{1}{8} \ln (| \tan (8 x) - 6 |) + C$