Calculus Examples

$f (x) = 8 \sec (x) (\sec (x) - 4 \tan (x))$

Step 1

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 2

Set up the integral to solve.

$F (x) = \int 8 \sec (x) (\sec (x) - 4 \tan (x)) d x$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\int 8 \sec (x) \sec (x) + 8 \sec (x) (- 4 \tan (x)) d x$

Step 3.2

Multiply $8 \sec (x) \sec (x)$ .

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Step 3.2.1

Raise $\sec (x)$ to the power of $1$ .

$\int 8 (\sec^{1} (x) \sec (x)) + 8 \sec (x) (- 4 \tan (x)) d x$

Step 3.2.2

Raise $\sec (x)$ to the power of $1$ .

$\int 8 (\sec^{1} (x) \sec^{1} (x)) + 8 \sec (x) (- 4 \tan (x)) d x$

Step 3.2.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 8 {\sec (x)}^{1 + 1} + 8 \sec (x) (- 4 \tan (x)) d x$

Step 3.2.4

Add $1$ and $1$ .

$\int 8 \sec^{2} (x) + 8 \sec (x) (- 4 \tan (x)) d x$

Step 3.3

Multiply $- 4$ by $8$ .

$\int 8 \sec^{2} (x) - 32 \sec (x) \tan (x) d x$

Step 4

Split the single integral into multiple integrals.

$\int 8 \sec^{2} (x) d x + \int - 32 \sec (x) \tan (x) d x$

Step 5

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$8 \int \sec^{2} (x) d x + \int - 32 \sec (x) \tan (x) d x$

Step 6

Since the derivative of $\tan (x)$ is $\sec^{2} (x)$ , the integral of $\sec^{2} (x)$ is $\tan (x)$ .

$8 (\tan (x) + C) + \int - 32 \sec (x) \tan (x) d x$

Step 7

Since $- 32$ is constant with respect to $x$ , move $- 32$ out of the integral.

$8 (\tan (x) + C) - 32 \int \sec (x) \tan (x) d x$

Step 8

Since the derivative of $\sec (x)$ is $\sec (x) \tan (x)$ , the integral of $\sec (x) \tan (x)$ is $\sec (x)$ .

$8 (\tan (x) + C) - 32 (\sec (x) + C)$

Step 9

Simplify.

$8 \tan (x) - 32 \sec (x) + C$

Step 10

The answer is the antiderivative of the function $f (x) = 8 \sec (x) (\sec (x) - 4 \tan (x))$ .

$F (x) =$ $8 \tan (x) - 32 \sec (x) + C$