Calculus Examples

$\tan^{5} (x)$

Step 1

Write $\tan^{5} (x)$ as a function.

$f (x) = \tan^{5} (x)$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

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

Step 4

Factor out $\tan^{4} (x)$ .

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

Step 5

Simplify with factoring out.

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Step 5.1

Factor $2$ out of $4$ .

$\int {\tan (x)}^{2 (2)} \tan (x) d x$

Step 5.2

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

$\int {(\tan^{2} (x))}^{2} \tan (x) d x$

Step 6

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

$\int {(- 1 + \sec^{2} (x))}^{2} \tan (x) d x$

Step 7

Use the Binomial Theorem.

$\int ({(- 1)}^{2} + 2 \cdot - 1 \sec^{2} (x) + {(\sec^{2} (x))}^{2}) \tan (x) d x$

Step 8

Simplify by multiplying through.

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Step 8.1

Apply the distributive property.

$\int {(- 1)}^{2} \tan (x) + 2 \cdot - 1 \sec^{2} (x) \tan (x) + {(\sec^{2} (x))}^{2} \tan (x) d x$

Step 8.2

Simplify.

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Step 8.2.1

Raise $- 1$ to the power of $2$ .

$\int 1 \tan (x) + 2 \cdot - 1 \sec^{2} (x) \tan (x) + {(\sec^{2} (x))}^{2} \tan (x) d x$

Step 8.2.2

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

$\int \tan (x) + 2 \cdot - 1 \sec^{2} (x) \tan (x) + {(\sec^{2} (x))}^{2} \tan (x) d x$

Step 8.2.3

Multiply $2$ by $- 1$ .

$\int \tan (x) - 2 \sec^{2} (x) \tan (x) + {(\sec^{2} (x))}^{2} \tan (x) d x$

Step 8.2.4

Multiply the exponents in ${(\sec^{2} (x))}^{2}$ .

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Step 8.2.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int \tan (x) - 2 \sec^{2} (x) \tan (x) + {\sec (x)}^{2 \cdot 2} \tan (x) d x$

Step 8.2.4.2

Multiply $2$ by $2$ .

$\int \tan (x) - 2 \sec^{2} (x) \tan (x) + \sec^{4} (x) \tan (x) d x$

Step 9

Split the single integral into multiple integrals.

$\int \tan (x) d x + \int - 2 \sec^{2} (x) \tan (x) d x + \int \sec^{4} (x) \tan (x) d x$

Step 10

The integral of $\tan (x)$ with respect to $x$ is $\ln (| \sec (x) |)$ .

$\ln (| \sec (x) |) + C + \int - 2 \sec^{2} (x) \tan (x) d x + \int \sec^{4} (x) \tan (x) d x$

Step 11

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$\ln (| \sec (x) |) + C - 2 \int \sec^{2} (x) \tan (x) d x + \int \sec^{4} (x) \tan (x) d x$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Differentiate $\sec (x)$ .

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

Step 12.1.2

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

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

Step 12.2

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

$\ln (| \sec (x) |) + C - 2 \int u_{1} d u_{1} + \int \sec^{4} (x) \tan (x) d x$

Step 13

By the Power Rule, the integral of $u_{1}$ with respect to $u_{1}$ is $\frac{1}{2} {u_{1}}^{2}$ .

$\ln (| \sec (x) |) + C - 2 (\frac{1}{2} {u_{1}}^{2} + C) + \int \sec^{4} (x) \tan (x) d x$

Step 14

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

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Step 14.1

Let $u_{2} = \sec (x)$ . Find $\frac{d u_{2}}{d x}$ .

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Step 14.1.1

Differentiate $\sec (x)$ .

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

Step 14.1.2

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

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

Step 14.2

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

$\ln (| \sec (x) |) + C - 2 (\frac{1}{2} {u_{1}}^{2} + C) + \int {u_{2}}^{3} d u_{2}$

Step 15

By the Power Rule, the integral of ${u_{2}}^{3}$ with respect to $u_{2}$ is $\frac{1}{4} {u_{2}}^{4}$ .

$\ln (| \sec (x) |) + C - 2 (\frac{1}{2} {u_{1}}^{2} + C) + \frac{1}{4} {u_{2}}^{4} + C$

Step 16

Simplify.

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Step 16.1

Combine $\frac{1}{2}$ and ${u_{1}}^{2}$ .

$\ln (| \sec (x) |) + C - 2 (\frac{{u_{1}}^{2}}{2} + C) + \frac{1}{4} {u_{2}}^{4} + C$

Step 16.2

Simplify.

$\ln (| \sec (x) |) - {u_{1}}^{2} + \frac{1}{4} {u_{2}}^{4} + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $\sec (x)$ .

$\ln (| \sec (x) |) - \sec^{2} (x) + \frac{1}{4} {u_{2}}^{4} + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $\sec (x)$ .

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

Step 18

The answer is the antiderivative of the function $f (x) = \tan^{5} (x)$ .

$F (x) =$ $\ln (| \sec (x) |) - \sec^{2} (x) + \frac{1}{4} \sec^{4} (x) + C$