1. ## Integration---I'm stuck!

$\int \tan^4\theta d\theta$
I know you're supposed to use the identity $\tan^2\theta+1=\sec^2\theta$
so, $\int \tan^4\theta d\theta=\int (\sec^2\theta-1)^2 d\theta$
now I don't know what to do next. Can anyone give me some pointers to get moving?
Thanks

2. Originally Posted by arze
$\int \tan^4\theta d\theta$
I know you're supposed to use the identity $\tan^2\theta+1=\sec^2\theta$
so, $\int \tan^4\theta d\theta=\int (\sec^2\theta-1)^2 d\theta$
now I don't know what to do next. Can anyone give me some pointers to get moving?
Thanks
$\tan^4{\theta} = \tan^2{\theta}\cdot\tan^2{\theta}$

$= \tan^2{\theta}(\sec^2{\theta} - 1)$

$= \tan^2{\theta}\sec^2{\theta} - \tan^2{\theta}$

$= \tan^2{\theta}\sec^2{\theta} - (\sec^2{\theta} - 1)$

$= \tan^2{\theta}\sec^2{\theta} - \sec^2{\theta} + 1$.

Therefore

$\int{\tan^4{\theta}\,d\theta} = \int{\tan^2{\theta}\sec^2{\theta}\,d\theta} - \int{\sec^2{\theta}\,d\theta} + \int{1\,d\theta}$

Let $u = \tan{\theta}$ so that $\frac{du}{d\theta} = \sec^2{\theta}$

So the integral becomes

$\int{u^2\,du} - \int{\sec^2{\theta}\,d\theta} + \int{1\,d\theta}$

$= \frac{1}{3}u^3 - \tan{\theta} + \theta + C$

$= \frac{1}{3}\tan^3{\theta} - \tan{\theta} + \theta + C$.