1. ## Trig Substitution

The problem is

Integrate sqrt(9x^2-25)/x^3 dx

To solve for this would I use a right triangle like so

I also think you might need to use a sec identity where the sqrt(x^2-a^2)=asec(theta) but I'm not sure how to set that up.

2. ## Re: Trig Substitution

The question is not clear as to what is given and what do you want to find out.

3. ## Re: Trig Substitution

Sorry, forgot to say that needs to be integrated.

4. ## Re: Trig Substitution

Originally Posted by goku900
The problem is

Integrate sqrt(9x^2-25)/x^3 dx

To solve for this would I use a right triangle like so

I also think you might need to use a sec identity where the sqrt(x^2-a^2)=asec(theta) but I'm not sure how to set that up.
It might be easier to use a Hyperbolic Substitution rather than a Trigonometric. Recall that $\displaystyle \displaystyle \cosh^2{(x)} - \sinh^2{(x)} = 1 \implies \cosh^2{(x)} - 1 = \sinh^2{(x)}$. Also $\displaystyle \displaystyle \frac{d}{dx} \left[ \cosh{(x)} \right] = \sinh{(x)}$.

So in this case, a substitution $\displaystyle \displaystyle x = \frac{5}{3}\cosh{(x)} \implies dx = \frac{5}{3}\sinh{(x)}\,dx$ is appropriate.

5. ## Re: Trig Substitution

You should have 3x on the hypotenuse.

So your substitution is $\displaystyle \frac{5}{3x}=\sin{\theta}$ or $\displaystyle \frac{1}{x}=\frac{3}{5}\sin{\theta}$ and then $\displaystyle \cot{\theta}=\frac{\sqrt{9x^2-25}}{5}$ or $\displaystyle 5\cot{\theta}=\sqrt{9x^2-25}$.

- Hollywood