# Integration With Trig Substitution Calc II

1. Feb 18, 2010

### Wm_Davies

1. The problem statement, all variables and given/known data

$$\int \frac{\sqrt{196 x^2-144}}{x} dx$$

2. Relevant equations

3. The attempt at a solution

I first rewrote the integral.....
$$\int \frac{\sqrt{(14x)^2-12^2}}{x} dx$$
Then I let.......
14x=12sec$$\theta$$
thus....
x=6/7sec$$\theta$$
dx=6/7sec$$\theta tan \theta d \theta$$

My progression of this problem is as follows:

$$\int \frac{\sqrt{(144(sec^2 \theta -1)}}{6/7sec \theta} d \theta$$
$$12 \int \frac{tan \theta}{6/7sec \theta}(6/7sec \theta tan \theta) d \theta$$
$$12 \int tan^2 \theta d \theta$$
$$12 \int sec^2 - 1 \theta d \theta$$
$$12 tan \theta$$

I am not sure how to finish the problem after that. I am also not sure I did that right since I am very new at this, but I have a feeling everything I did was right.

2. Feb 18, 2010

### LCKurtz

I didn't check all your steps, but didn't you forget to integrate the -θ in the last integral?

Now draw a little right triangle in the first quadrant with hypotenuse 7x, x-side of 6 and y-side of $\sqrt{(49x^2-36)}$ and polar angle θ. You should be able to read everything off this triangle to get your answer back in terms of x.

3. Feb 18, 2010

### Wm_Davies

Sorry the last integral was a typo it should be....

$$12 \int sec^2 \theta - 1 d \theta$$

Okay so after I fill this in the answer should be....

$$\frac{\sqrt{49x^2-36}}{6}+c$$

4. Feb 18, 2010

### Wm_Davies

That answer is wrong. I am not sure what mistake I made.

5. Feb 19, 2010

### Wm_Davies

I forgot to multiply it by 12 so I would have had...

$$2 \sqrt{49x^2 - 36} + c$$

which is still wrong

I checked Wolfram Alpha and they gave me the answer as.....

$$2 \sqrt{49 x^2 - 36} + 12 tan^-1\left[\frac {6}{\sqrt{49 x^2 -36}}\right] + c$$

So I now know that I am close, but after checking it with the steps in Wolfram Alpha I am still not sure where I went wrong.

Last edited: Feb 19, 2010
6. Feb 19, 2010

### Bohrok

What is ∫tan2θ dθ? There's really no need to change tan2θ to sec2θ - 1 that I can see.
Once you find that integral, make sure you substitute for x correctly.

7. Feb 19, 2010

### vela

Staff Emeritus
You forgot to integrate the 1 in the integrand.

8. Feb 20, 2010

### Wm_Davies

Wow, I did forget to do that! Thanks for noticing that. I got it now.