# How to integrate $\int_{}^{}{\frac{\sin ^{3}\theta }{\cos ^{6}\theta }d\theta }$?

How to integrate $\int_{}^{}{\frac{\sin ^{3}\theta }{\cos ^{6}\theta }d\theta }$? This is kind of homework,and I have no idea where to start.

-

You can rewrite it as $$\int\tan^3\theta \sec^3\theta d\theta.$$ Note that $d(\sec\theta)=\tan\theta\sec\theta d\theta$ and $\tan^2\theta=\sec^2\theta-1$, we have $$\int\tan^3\theta \sec^3\theta d\theta=\int \tan^2\theta \sec^2\theta d(\sec\theta) =\int (\sec^2\theta-1) \sec^2\theta d(\sec\theta).$$ Now if we let $t=\sec\theta$, then....I leave the remaining parts to you.

-
Thanks,I got it! –  Ave Maleficum Jan 25 '13 at 11:45

Or We can also Calculate it Using Trig. Substitution

$\displaystyle \int\frac{\sin^3 \theta}{\cos^6 \theta}d\theta = \int\frac{1-\cos^2 \theta}{\cos^6 \theta} \times \sin \theta d\theta$

Let $\cos \theta = t$ and $\sin \theta d\theta = -dt$ So Integral is

$\displaystyle \int\frac{t^2-1}{t^6}dt = \int t^{-4}dt-\int t^{-6}dt$

$\displaystyle = -\frac{1}{3}t^{-3}+\frac{1}{5}t^{-5}+\mathbb{C}$

$\displaystyle = -\frac{1}{3}(\cos t )^{-3}+\frac{1}{5}(\cos t) ^{-5}+\mathbb{C}$

-
Is there a reason you have used $\mathbb C$ to denote the constant? It seems rather peculiar to me. –  user50407 Jan 25 '13 at 22:23