# Math Help - Trigonometric Substitution Integral

1. ## Trigonometric Substitution Integral

I'm having trouble on one specific step of this integral:

Original Equation: $\int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx$

where $x = \sqrt{5} tan\Theta$
$dx = \sqrt{5} sec^2\Theta d\Theta$

= $\int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta$

Now I don't understand how to get to the next step of:

$-\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta$

because I get here and I know it's not right:

$-5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta$

Thanks for the help in advance.

2. Originally Posted by cynlix
I'm having trouble on one specific step of this integral:

Original Equation: $\int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx$

where $x = \sqrt{5} tan\Theta$
$dx = \sqrt{5} sec^2\Theta d\Theta$

= $\int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta$

Now I don't understand how to get to the next step of:

$-\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta$

because I get here and I know it's not right:

$-5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta$

Thanks for the help in advance.
You can just use a $u$ substitution.

Let $u = x^2 + 5$ so that $\frac{du}{dx} = 2x$.

The integral becomes:

$\int{\frac{-5x}{(x^2 + 5)^{\frac{3}{2}}}\,dx} = -\frac{5}{2}\int{\frac{2x}{(x^2 + 5)^{\frac{3}{2}}}\,dx}$

$= -\frac{5}{2}\int{\frac{1}{u^{\frac{3}{2}}}\,\frac{d u}{dx}\,dx}$

$= -\frac{5}{2}\int{u^{-\frac{3}{2}}\,du}$.

You should be able to go from here...

3. thank you for that insight, however I need to use the trigonometric substitution for this homework problem.

4. Originally Posted by cynlix
I'm having trouble on one specific step of this integral:

Original Equation: $\int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx$

where $x = \sqrt{5} tan\Theta$
$dx = \sqrt{5} sec^2\Theta d\Theta$

= $\int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta$

Now I don't understand how to get to the next step of:

$-\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta$

because I get here and I know it's not right:

$-5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta$

Thanks for the help in advance.
You also need to replace the $x$ in your numerator with $\sqrt{5}\tan{\theta}$.