Calculus Examples

$\int \tan^{4} (5 x) d x$

Step 1

Let $u_{1} = 5 x$ . Then $d u_{1} = 5 d x$ , so $\frac{1}{5} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

Let $u_{1} = 5 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.1.1

Differentiate $5 x$ .

$\frac{d}{d x} [5 x]$

Step 1.1.2

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$5 \frac{d}{d x} [x]$

Step 1.1.3

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

$5 \cdot 1$

Step 1.1.4

Multiply $5$ by $1$ .

$5$

Step 1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \tan^{4} (u_{1}) \frac{1}{5} d u_{1}$

Step 2

Combine $\tan^{4} (u_{1})$ and $\frac{1}{5}$ .

$\int \frac{\tan^{4} (u_{1})}{5} d u_{1}$

Step 3

Since $\frac{1}{5}$ is constant with respect to $u_{1}$ , move $\frac{1}{5}$ out of the integral.

$\frac{1}{5} \int \tan^{4} (u_{1}) d u_{1}$

Step 4

Simplify with factoring out.

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Step 4.1

Factor $2$ out of $4$ .

$\frac{1}{5} \int {\tan (u_{1})}^{2 (2)} d u_{1}$

Step 4.2

Rewrite ${\tan (u_{1})}^{2 (2)}$ as exponentiation.

$\frac{1}{5} \int {(\tan^{2} (u_{1}))}^{2} d u_{1}$

Step 5

Using the Pythagorean Identity, rewrite $\tan^{2} (u_{1})$ as $- 1 + \sec^{2} (u_{1})$ .

$\frac{1}{5} \int {(- 1 + \sec^{2} (u_{1}))}^{2} d u_{1}$

Step 6

Simplify.

$\frac{1}{5} \int 1 - 2 \sec^{2} (u) + \sec^{4} (u) d u_{1}$

Step 7

Apply the constant rule.

$\frac{1}{5} ((1 - 2 \sec^{2} (u) + \sec^{4} (u)) u_{1} + C)$

Step 8

Simplify the answer.

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Step 8.1

Rewrite $\frac{1}{5} ((1 - 2 \sec^{2} (u) + \sec^{4} (u)) u_{1} + C)$ as $\frac{1}{5} (u_{1} - 2 \sec^{2} (u) u_{1} + \sec^{4} (u) u_{1}) + C$ .

$\frac{1}{5} (u_{1} - 2 \sec^{2} (u) u_{1} + \sec^{4} (u) u_{1}) + C$

Step 8.2

Replace all occurrences of $u_{1}$ with $5 x$ .

$\frac{1}{5} (5 x - 2 \sec^{2} (u) (5 x) + \sec^{4} (u) (5 x)) + C$

Step 8.3

Remove parentheses.

$\frac{1}{5} (5 x - 2 \sec^{2} (u) \cdot 5 x + \sec^{4} (u) \cdot 5 x) + C$