Calculus Examples

$\sec^{3} (θ)$

Step 1

Write $\sec^{3} (θ)$ as a function.

$f (θ) = \sec^{3} (θ)$

Step 2

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

$F (θ) = \int f (θ) d θ$

Step 3

Set up the integral to solve.

$F (θ) = \int \sec^{3} (θ) d θ$

Step 4

Factor $\sec (θ)$ out of $\sec^{3} (θ)$ .

$\int \sec (θ) \sec^{2} (θ) d θ$

Step 5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \sec (θ)$ and $d v = \sec^{2} (θ)$ .

$\sec (θ) \tan (θ) - \int \tan (θ) (\sec (θ) \tan (θ)) d θ$

Step 6

Raise $\tan (θ)$ to the power of $1$ .

$\sec (θ) \tan (θ) - \int \tan^{1} (θ) \tan (θ) \sec (θ) d θ$

Step 7

Raise $\tan (θ)$ to the power of $1$ .

$\sec (θ) \tan (θ) - \int \tan^{1} (θ) \tan^{1} (θ) \sec (θ) d θ$

Step 8

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

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

Step 9

Simplify the expression.

Tap for more steps...

Step 9.1

Add $1$ and $1$ .

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

Step 9.2

Reorder $\tan^{2} (θ)$ and $\sec (θ)$ .

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

Step 10

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

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

Step 11

Simplify by multiplying through.

Tap for more steps...

Step 11.1

Rewrite the exponentiation as a product.

$\sec (θ) \tan (θ) - \int \sec (θ) (- 1 + \sec (θ) \sec (θ)) d θ$

Step 11.2

Apply the distributive property.

$\sec (θ) \tan (θ) - \int \sec (θ) \cdot - 1 + \sec (θ) (\sec (θ) \sec (θ)) d θ$

Step 11.3

Reorder $\sec (θ)$ and $- 1$ .

$\sec (θ) \tan (θ) - \int - 1 \cdot \sec (θ) + \sec (θ) (\sec (θ) \sec (θ)) d θ$

Step 12

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

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

Step 13

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

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

Step 14

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

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

Step 15

Add $1$ and $1$ .

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

Step 16

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

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

Step 17

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

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

Step 18

Add $2$ and $1$ .

$\sec (θ) \tan (θ) - \int - 1 \sec (θ) + \sec^{3} (θ) d θ$

Step 19

Split the single integral into multiple integrals.

$\sec (θ) \tan (θ) - (\int - 1 \sec (θ) d θ + \int \sec^{3} (θ) d θ)$

Step 20

Since $- 1$ is constant with respect to $θ$ , move $- 1$ out of the integral.

$\sec (θ) \tan (θ) - (- \int \sec (θ) d θ + \int \sec^{3} (θ) d θ)$

Step 21

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

$\sec (θ) \tan (θ) - (- (\ln (| \sec (θ) + \tan (θ) |) + C) + \int \sec^{3} (θ) d θ)$

Step 22

Simplify by multiplying through.

Tap for more steps...

Step 22.1

Apply the distributive property.

$\sec (θ) \tan (θ) - - (\ln (| \sec (θ) + \tan (θ) |) + C) - \int \sec^{3} (θ) d θ$

Step 22.2

Multiply $- 1$ by $- 1$ .

$\sec (θ) \tan (θ) + 1 (\ln (| \sec (θ) + \tan (θ) |) + C) - \int \sec^{3} (θ) d θ$

Step 23

Solving for $\int \sec^{3} (θ) d θ$ , we find that $\int \sec^{3} (θ) d θ$ = $\frac{\sec (θ) \tan (θ) + 1 (\ln (| \sec (θ) + \tan (θ) |) + C)}{2}$ .

$\frac{\sec (θ) \tan (θ) + 1 (\ln (| \sec (θ) + \tan (θ) |) + C)}{2} + C$

Step 24

Multiply $\ln (| \sec (θ) + \tan (θ) |) + C$ by $1$ .

$\frac{\sec (θ) \tan (θ) + \ln (| \sec (θ) + \tan (θ) |) + C}{2} + C$

Step 25

Simplify.

$\frac{1}{2} (\sec (θ) \tan (θ) + \ln (| \sec (θ) + \tan (θ) |)) + C$

Step 26

The answer is the antiderivative of the function $f (θ) = \sec^{3} (θ)$ .

$F (θ) =$ $\frac{1}{2} (\sec (θ) \tan (θ) + \ln (| \sec (θ) + \tan (θ) |)) + C$