Integration by Trig Substitution $u=a \tan(\theta)$ I am currently working on a problem that deals with trig sub to integrate. The worksheet that I have gotten it off of says to reduce it into sine and cosine functions and use the Table of Integrals given:

Here is my work that I have done so far:

From here on out I do not know what to do. It looks nothing like anything in the table of integrals given so I do not know how to move foward!
Thanks in advance for your help!!
 A: With $x=\sin\theta$ this integral is $\frac{1}{81}\int\frac{1-x^2}{x^2}dx$. You can do the rest.
A: Alternatively using the identity $\cot^2(x)+1=\csc^2(x)$ and that $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$ we have
$$\frac{1}{81}\int\frac{1}{\sec(\theta)\tan^2(\theta)}d\theta=\frac{1}{81}\int\cos(\theta)\cot^{2}(\theta)d\theta$$
$$=\frac{1}{81}\int(\cos(\theta)(\csc^2(\theta)-1)d\theta=\frac{1}{81}\int(\csc(\theta)\cot(\theta)-\cos(\theta))d\theta$$
$$=\frac{1}{81}\big[\int\cot(\theta)\csc(\theta)d\theta-\int \cos(\theta)d\theta\big]$$
$$=\frac{1}{81}\big[-\csc(\theta)-\sin(\theta)\big]+C$$
$$=\frac{1}{81}\big[-\sin(\theta)-\frac{1}{\sin(\theta)}\big]+C$$
$$=\frac{1}{81}\big[\frac{t}{3\sqrt{1+\frac{t^2}{9}}}-\frac{1}{\frac{t}{3\sqrt{1+\frac{t^2}{9}}}}\big]+C$$
$$=-\frac{t}{243\sqrt{1+\frac{t^2}{9}}}-\frac{\sqrt{1+\frac{t^2}{9}}}{27t}+C$$
$$=-\frac{2t^2+9}{81t\sqrt{t^2+9}}+C$$
A: You have
$$\int \sin^{-2}\theta \cos^3\theta\, \mathrm d\theta. $$
This fits formula number 23 on your worksheet with $m=-2$ and $n = 3.$
