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
$\displaystyle \frac{dy}{dx} = 4\sec^2{x}\tan^3{x}$.
$\displaystyle \int \tan^4{x} \, dx$
$\displaystyle \int \tan^2{x}(\sec^2{x} - 1) \, dx$
$\displaystyle \int \tan^2{x}\sec^2{x} - \tan^2{x} \, dx$
$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \tan^2{x} \, dx$
$\displaystyle \int \tan^2{x}\sec^2{x} \, dx - \int \sec^2{x} - 1 \, dx$
first integral ... $\displaystyle u =tan{x}$ ... $\displaystyle du = \sec^2{x} \, dx$
$\displaystyle \frac{\tan^3{x}}{3} - \tan{x} + x + C$