# Thread: [SOLVED] trig substitution integral

1. ## [SOLVED] trig substitution integral

Evaluate the integral:

$\int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

Here is what I am getting in my first step:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (sec\theta - 1)} d\theta$

Is that the right thing to do with the $x^3$ in the denominator?

After that, assuming it is correct, I simplified the denominator:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^4 - (sec\theta)^3} d\theta$

Is this correct to this point? If yes, where do I go from here?

THanks!!

2. Originally Posted by mollymcf2009
Evaluate the integral:

$\int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

Here is what I am getting in my first step:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot ({\color{red}sec\theta - 1})} d\theta$

Is that the right thing to do with the $x^3$ in the denominator? Yes
.
.
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what is in red is incorrect. can you fix it?

3. Also, change the limits of integration: $x = \sec \theta$

• $\sqrt{2} = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{4}$
• ${\color{white}.} \ 2 = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{3}$

4. Originally Posted by Jhevon
what is in red is incorrect. can you fix it?
[quote=mollymcf2009;264793]Evaluate the integral:

$\int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

Here is what I am getting in my first step:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (\sqrt{sec^2\theta - 1})} d\theta$

Is that the right thing to do with the $x^3$ in the denominator?

Better? What is my next step?

5. Originally Posted by o_O
Also, change the limits of integration: $x = \sec \theta$

• $\sqrt{2} = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{4}$
• ${\color{white}.} \ 2 = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{3}$
But I need my answer in terms of x for webassign, should I still change them?

6. [quote=mollymcf2009;264800]
Originally Posted by mollymcf2009
Evaluate the integral:

$\int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

Here is what I am getting in my first step:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (\sqrt{sec^2\theta - 1})} d\theta$

Is that the right thing to do with the $x^3$ in the denominator?

Better? What is my next step?
note that $\sec^2 \theta - 1 = \tan^2 \theta$

with that in mind, simplify the integrand before you integrate

Originally Posted by mollymcf2009
But I need my answer in terms of x for webassign, should I still change them?
your answer will be a constant... either you use the x-limits with x or the u-limits with u. your choice. but you cannot write the x-limits on the u-integral

7. It's a definite integral. You'll be entering a numerical value. So all you need here is consistency.

________________

$\int \frac{\sec \theta \tan \theta d \theta}{\sec^3 \theta \sqrt{\sec^2 \theta - 1}}$

Cancel the $\sec \theta$ from the numerator with one in the denominator: $\int \frac{\tan \theta d \theta}{\sec^2 \theta\sqrt{\sec^2 \theta - 1}}$

Use the identity $\sec^2 \theta - 1 = \tan^2 \theta$ to get: $\int \frac{\tan \theta d \theta}{\sec^2 \theta \cdot \sqrt{\tan^2 \theta}}$

etc. etc.

8. Just want to double check before I continue:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{tan^2\theta \cdot sec\theta \cdot \sqrt{tan^2\theta}} d\theta$

9. Originally Posted by mollymcf2009
Just want to double check before I continue:

$\int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{tan^2\theta \cdot sec\theta \cdot \sqrt{tan^2\theta}} d\theta$

10. Sorry! I didn't see it until after I posted. I've got it now. Thanks everyone for your help

11. Originally Posted by o_O
It's a definite integral. You'll be entering a numerical value. So all you need here is consistency.

________________

$\int \frac{\sec \theta \tan \theta d \theta}{\sec^3 \theta \sqrt{\sec^2 \theta - 1}}$

Cancel the $\sec \theta$ from the numerator with one in the denominator: $\int \frac{\tan \theta d \theta}{\sec^2 \theta\sqrt{\sec^2 \theta - 1}}$

Use the identity $\sec^2 \theta - 1 = \tan^2 \theta$ to get: $\int \frac{\tan \theta d \theta}{\sec^2 \theta \cdot \sqrt{\tan^2 \theta}}$

etc. etc.
This is driving me crazy! I have never had this much trouble grasping a concept.

I'm gun shy about canceling for some reason. I feel like I need to do a u substitution instead. Can I cancel out $tan\theta$ in top and bottom since the bottom is sqrt of $tan^2$ ?? Leaving me with $\frac{1}{sec^2\theta}$ or am I supposed to use a u sub?

12. Originally Posted by mollymcf2009
This is driving me crazy! I have never had this much trouble grasping a concept.

I'm gun shy about canceling for some reason. I feel like I need to do a u substitution instead. Can I cancel out $tan\theta$ in top and bottom since the bottom is sqrt of $tan^2$ ?? Leaving me with $\frac{1}{sec^2\theta}$
yes, this is what you should do.

now, note that $\frac 1{\sec^2 x} = \cos^2 x$

13. Jhevon! Thank you so much for your patience and help! I greatly appreciate it!! Molly