Calculus Examples

$\frac{\sec^{3} (θ) \tan (θ)}{1 + \tan^{2} (θ)}$

Step 1

Write $\frac{\sec^{3} (θ) \tan (θ)}{1 + \tan^{2} (θ)}$ as a function.

$f (θ) = \frac{\sec^{3} (θ) \tan (θ)}{1 + \tan^{2} (θ)}$

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.

$F (θ) = \int \frac{\sec^{3} (θ) \tan (θ)}{1 + \tan^{2} (θ)} d θ$

Step 4

Rearrange terms.

$\int \frac{\sec^{3} (θ) \tan (θ)}{\tan^{2} (θ) + 1} d θ$

Step 5

Apply pythagorean identity.

$\int \frac{\sec^{3} (θ) \tan (θ)}{\sec^{2} (θ)} d θ$

Step 6

Cancel the common factor of $\sec^{3} (θ)$ and $\sec^{2} (θ)$ .

Tap for more steps...

Step 6.1

Factor $\sec^{2} (θ)$ out of $\sec^{3} (θ) \tan (θ)$ .

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

Step 6.2

Cancel the common factors.

Tap for more steps...

Step 6.2.1

Multiply by $1$ .

$\int \frac{\sec^{2} (θ) (\sec (θ) \tan (θ))}{\sec^{2} (θ) \cdot 1} d θ$

Step 6.2.2

Cancel the common factor.

$\int \frac{\sec^{2} (θ) (\sec (θ) \tan (θ))}{\sec^{2} (θ) \cdot 1} d θ$

Step 6.2.3

Rewrite the expression.

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

Step 6.2.4

Divide $\sec (θ) \tan (θ)$ by $1$ .

$\int \sec (θ) \tan (θ) d θ$

Step 7

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

$\sec (θ) + C$

Step 8

The answer is the antiderivative of the function $f (θ) = \frac{\sec^{3} (θ) \tan (θ)}{1 + \tan^{2} (θ)}$ .

$F (θ) =$ $\sec (θ) + C$