Trgnometric integrals

**radioheadfan** · October 24th 2009, 05:40 PM

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?

**Prove It** · October 24th 2009, 05:43 PM

Originally Posted by radioheadfan

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.

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

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

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

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

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

Therefore

$\frac{dy}{dx} = 4\sec^2{x}\tan^3{x}$ .

**skeeter** · October 24th 2009, 05:57 PM

Originally Posted by radioheadfan

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?

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

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

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

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

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

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

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

**Prove It** · October 24th 2009, 06:13 PM

Originally Posted by skeeter

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

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

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

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

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

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

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

Oops, I read that it had to be differentiated. LOL.

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