1. ## Trigonometric Substitution Integrals

How do u integrate this function:

$\displaystyle \int x^2 {\sqrt{x^2-16}}\,dx\,$

I got it down to this but now i cant solve this integral:

$\displaystyle 256\int \tan^2\theta\,sec^3\theta\, d\theta\,$

Which identity do i sub in....?

2. $\displaystyle \displaystyle\int\tan^2x\sec^3xdx=\int\frac{\sin^2 x}{\cos^5x}dx=\int\frac{\sin^2x\cos x}{\cos^6x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^3}\cos xdx$
Now $\displaystyle \sin x=t\Rightarrow\int\displaystyle\frac{t^2}{(1-t^2)^3}dt$
and this is a rational integral.

3. Originally Posted by red_dog
$\displaystyle \displaystyle\int\tan^2x\sec^3xdx=\int\frac{\sin^2 x}{\cos^5x}dx=\int\frac{\sin^2x\cos x}{\cos^6x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^3}\cos xdx$
Now $\displaystyle \sin x=t\Rightarrow\int\displaystyle\frac{t^2}{(1-t^2)^3}dt$
and this is a rational integral.
How do u solve this function...it's almost as hard as the first

4. I may be mistaken ,but I believe that's 128 and not 256.

Anyway, $\displaystyle \int{tan^{2}(x)sec^{3}(x)}dx$

Using identities, rewrite as:

$\displaystyle \int{(sec^{2}(x)-1)sec^{3}(x)}dx$

=$\displaystyle \int{sec^{5}(x)}dx-\int{sec^{3}(x)}dx$

Now, you can use the reduction formula for sec:

$\displaystyle \int{sec^{m}(x)}dx=\frac{1}{m-1}sec^{m-2}(x)tan(x)+\frac{m-2}{m-1}\int{sec^{m-2}(x)}dx$

Also, you may want it in terms of x. So, use the fact that $\displaystyle {\theta}=sec^{-1}(\frac{x}{4})$ and make the sub.