# Trgnometric integrals

• Oct 24th 2009, 05:40 PM
Trgnometric integrals
How would you evaluate this indefinite integral?
tan^4x dx

I began with Seperating it into two tans. Both to the 2nd power.
Then I got lost. How do I get to my answer?
• Oct 24th 2009, 05:43 PM
Prove It
Quote:

How would you evaluate this indefinite integral?
tan^4x dx

I began with Seperating it into two tans. Both to the 2nd power.
Then I got lost. How do I get to my answer?

Use the chain rule.

$\displaystyle y = \tan^4{x} = (\tan{x})^4$.

Let $\displaystyle u = \tan{x}$ so that $\displaystyle y = u^4$.

$\displaystyle \frac{du}{dx} = \sec^2{x}$

$\displaystyle \frac{dy}{du} = 4u^3$

$\displaystyle = 4\tan^3{x}$.

Therefore

$\displaystyle \frac{dy}{dx} = 4\sec^2{x}\tan^3{x}$.
• Oct 24th 2009, 05:57 PM
skeeter
Quote:

How would you evaluate this indefinite integral?
tan^4x dx

I began with Seperating it into two tans. Both to the 2nd power.
Then I got lost. How do I get to my answer?

$\displaystyle \int \tan^4{x} \, dx$

$\displaystyle \int \tan^2{x}(\sec^2{x} - 1) \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx$

first integral ... $\displaystyle u =tan{x}$ ... $\displaystyle du = \sec^2{x} \, dx$

$\displaystyle \frac{\tan^3{x}}{3} - \tan{x} + x + C$
• Oct 24th 2009, 06:13 PM
Prove It
Quote:

Originally Posted by skeeter
$\displaystyle \int \tan^4{x} \, dx$

$\displaystyle \int \tan^2{x}(\sec^2{x} - 1) \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx$

$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx$

first integral ... $\displaystyle u =tan{x}$ ... $\displaystyle du = \sec^2{x} \, dx$

$\displaystyle \frac{\tan^3{x}}{3} - \tan{x} + x + C$