Question Evaluate the integral of $\tan^3(x)\sec^6(x)$ and include the constant of integration.

Studdy Solution

STEP 1

Assumptions
1. We are given the integral of a trigonometric function to evaluate: $\int \tan^{3}(x) \sec^{6}(x) dx$
2. We will use trigonometric identities and integration techniques to solve the integral.
3. The constant of integration will be denoted as $C$ .

STEP 2

Recognize that $\tan^2(x) = \sec^2(x) - 1$ . This identity will be useful for simplifying the integral.

STEP 3

Rewrite $\tan^{3}(x)$ as $\tan(x) \cdot \tan^2(x)$ to use the trigonometric identity.
$\int \tan^{3}(x) \sec^{6}(x) dx = \int \tan(x) (\tan^2(x)) \sec^{6}(x) dx$

STEP 4

Apply the trigonometric identity $\tan^2(x) = \sec^2(x) - 1$ to the integral.
$\int \tan(x) (\tan^2(x)) \sec^{6}(x) dx = \int \tan(x) (\sec^2(x) - 1) \sec^{6}(x) dx$

STEP 5

Distribute $\sec^{6}(x)$ inside the parentheses.
$\int \tan(x) (\sec^2(x) - 1) \sec^{6}(x) dx = \int \tan(x) \sec^{8}(x) dx - \int \tan(x) \sec^{6}(x) dx$

STEP 6

We will solve the integrals separately. Let's start with the first integral $\int \tan(x) \sec^{8}(x) dx$ .

STEP 7

Use substitution to solve the first integral. Let $u = \sec(x)$ , which implies $du = \sec(x)\tan(x)dx$ .

STEP 8

Rewrite the first integral in terms of $u$ .
$\int \tan(x) \sec^{8}(x) dx = \int u^{8} du$

STEP 9

Integrate $u^{8}$ with respect to $u$ .
$\int u^{8} du = \frac{u^{9}}{9} + C_1$

STEP 10

Substitute back $\sec(x)$ for $u$ in the first integral.
$\frac{u^{9}}{9} + C_1 = \frac{\sec^{9}(x)}{9} + C_1$

STEP 11

Now, let's solve the second integral $\int \tan(x) \sec^{6}(x) dx$ .

STEP 12

Use substitution again. Let $v = \sec(x)$ , which implies $dv = \sec(x)\tan(x)dx$ .

STEP 13

Rewrite the second integral in terms of $v$ .
$\int \tan(x) \sec^{6}(x) dx = \int v^{6} dv$

STEP 14

Integrate $v^{6}$ with respect to $v$ .
$\int v^{6} dv = \frac{v^{7}}{7} + C_2$

STEP 15

Substitute back $\sec(x)$ for $v$ in the second integral.
$\frac{v^{7}}{7} + C_2 = \frac{\sec^{7}(x)}{7} + C_2$

STEP 16

Combine the results from the two integrals, remembering to subtract the second integral from the first.
$\frac{\sec^{9}(x)}{9} + C_1 - \left(\frac{\sec^{7}(x)}{7} + C_2\right)$

STEP 17

Combine the constants of integration $C_1$ and $C_2$ into a single constant $C$ .
$\frac{\sec^{9}(x)}{9} - \frac{\sec^{7}(x)}{7} + C$

STEP 18

Write the final answer.
The evaluated integral is: $\int \tan^{3}(x) \sec^{6}(x) dx = \frac{\sec^{9}(x)}{9} - \frac{\sec^{7}(x)}{7} + C$

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