# [SOLVED] trig substitution integral

• Feb 11th 2009, 08:09 PM
mollymcf2009
[SOLVED] trig substitution integral
Evaluate the integral:

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

Here is what I am getting in my first step:

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

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

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

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

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

THanks!!
• Feb 11th 2009, 08:11 PM
Jhevon
Quote:

Originally Posted by mollymcf2009
Evaluate the integral:

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

Here is what I am getting in my first step:

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

Is that the right thing to do with the $\displaystyle x^3$ in the denominator? Yes
.
.
.

what is in red is incorrect. can you fix it?
• Feb 11th 2009, 08:15 PM
o_O
Also, change the limits of integration: $\displaystyle x = \sec \theta$

• $\displaystyle \sqrt{2} = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{4}$
• $\displaystyle {\color{white}.} \ 2 = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{3}$
• Feb 11th 2009, 08:16 PM
mollymcf2009
Quote:

Originally Posted by Jhevon
what is in red is incorrect. can you fix it?

[quote=mollymcf2009;264793]Evaluate the integral:

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

Here is what I am getting in my first step:

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

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

Better? What is my next step?
• Feb 11th 2009, 08:18 PM
mollymcf2009
Quote:

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

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

But I need my answer in terms of x for webassign, should I still change them?
• Feb 11th 2009, 08:21 PM
Jhevon
[quote=mollymcf2009;264800]
Quote:

Originally Posted by mollymcf2009
Evaluate the integral:

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

Here is what I am getting in my first step:

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

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

Better? What is my next step?

note that $\displaystyle \sec^2 \theta - 1 = \tan^2 \theta$

with that in mind, simplify the integrand before you integrate

Quote:

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
• Feb 11th 2009, 08:22 PM
o_O
It's a definite integral. You'll be entering a numerical value. So all you need here is consistency.

________________

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

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

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

etc. etc.
• Feb 11th 2009, 08:36 PM
mollymcf2009
Just want to double check before I continue:

$\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{tan^2\theta \cdot sec\theta \cdot \sqrt{tan^2\theta}} d\theta$
• Feb 11th 2009, 08:39 PM
Jhevon
Quote:

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

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

• Feb 11th 2009, 08:41 PM
mollymcf2009
Sorry! I didn't see it until after I posted. I've got it now. Thanks everyone for your help
• Feb 11th 2009, 09:08 PM
mollymcf2009
Quote:

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

________________

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

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

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

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 $\displaystyle tan\theta$ in top and bottom since the bottom is sqrt of $\displaystyle tan^2$ ?? Leaving me with $\displaystyle \frac{1}{sec^2\theta}$ or am I supposed to use a u sub?
• Feb 11th 2009, 09:12 PM
Jhevon
Quote:

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 $\displaystyle tan\theta$ in top and bottom since the bottom is sqrt of $\displaystyle tan^2$ ?? Leaving me with $\displaystyle \frac{1}{sec^2\theta}$

yes, this is what you should do.

now, note that $\displaystyle \frac 1{\sec^2 x} = \cos^2 x$
• Feb 11th 2009, 09:15 PM
mollymcf2009
Jhevon! Thank you so much for your patience and help! I greatly appreciate it!! Molly