# Integration, Trig substitution

• Dec 12th 2012, 06:34 AM
Tutu
Integration, Trig substitution
Hi I tried these three questions, on trig substitution and I didn't get the answer for all three..I'm sorry I cannot show my whole working, latex is hard for me and I'm struggling trying my best to type the more 'important' parts in it. Hopefully it turns out fine..

1.) Integrate $\displaystyle \frac{1}{x^2\sqrt{16-x^2}}dx$
I had let x be $\displaystyle 4\sin\theta$, dx = $\displaystyle 4\cos\theta$ d(delta sign)
Subbing it in and simplifying, I got $\displaystyle \frac{1}{16\sin\theta}$ left to integrate..After integrating I got $\displaystyle \frac{-cot\theta}{16}+c$ and since x = $\displaystyle 4\sin\theta$, $\displaystyle \sin\theta = \frac{x}{4}$..
Substituting, I nearly got the answer,
My answer is $\displaystyle \frac{\-sqrt{16-x^2}}{4x} + c$ but it's wrong..

2.) Integrate $\displaystyle \frac{1}{x\(x^2+16)}dx$
I let x be $\displaystyle 4\tan\theta$, dx = $\displaystyle 4\sec^2\theta$ d(delta sign)
Subbing in and simplifying, I got to integrating $\displaystyle \frac{tan\theta}{16}$, used the substitution method to tackle this integration of $\displaystyle tan\theta$, got $\displaystyle \frac{1}{16}\ln|cos\theta|}+c$ but then I tried to put x in terms of delta, it didn't work out for I get really weird looking answers...!

3.) Integrate x^2(sqrt{4-x^2}), I had let x be $\displaystyle 4\sin\theta$, dx = $\displaystyle 2\cos\theta$ d(delta sign)
Subbing and simplifying, I was left with integrating $\displaystyle \16\cos\^2\theta\-\16\cos\^4\theta}dx$. Using the double angle expression, I simplified, integrated and got $\displaystyle 4\sin\2\theta\-\2\sin\4\theta + c$... continued with it but I'm getting nowhere!

Thank you so much for your time!
P.s. the trigo functions with values like 2 and 4 with that little ^ above them, sorry for my bad skills, it actually means that it is eg. (cos(x))^2 etc!
• Dec 12th 2012, 09:53 AM
ebaines
Re: Integration, Trig substitution
Quote:

Originally Posted by Tutu
After integrating I got $\displaystyle \frac{-cot\theta}{16}+c$ and since x = $\displaystyle 4\sin\theta$, $\displaystyle \sin\theta = \frac{x}{4}$..
Substituting, I nearly got the answer,
My answer is $\displaystyle \frac{\-sqrt{16-x^2}}{4x} + c$ but it's wrong..

Almost got it - a minor math error is all that's wrong. The denominator should be 16x, not 4x. If $\displaystyle \sin \theta = \frac x 4$, then $\displaystyle \cos \theta = \frac 1 4 \sqrt {16 - x^2}$ and $\displaystyle \frac {-cot \theta}{16} = - \frac {\frac 1 4 \sqrt {16 - x^2}}{16 \frac x 4} = \frac {- \sqrt {16-x^2}}{16x}$

Quote:

Originally Posted by Tutu
2.) ...Subbing in and simplifying, I got to integrating $\displaystyle \frac{tan\theta}{16}$

You should have $\displaystyle \frac {cot \theta} {16}$ here.

Quote:

Originally Posted by Tutu
3.) ...Using the double angle expression, I simplified, integrated and got $\displaystyle 4\sin\2\theta\-\2\sin\4\theta + c$... continued with it but I'm getting nowhere!

At this point you should have $\displaystyle 4\sin^2\theta-2\sin^4\theta + c$. To integrate $\displaystyle sin^2x$ use the identity $\displaystyle \sin^2 \theta = \frac { 1 - \cos(2 \theta)} 2$. And to integrate $\displaystyle sin^4 \theta$ treat is as $\displaystyle (sin^2 \theta) ^2$

$\displaystyle (sin^2 \theta)^2 = (\frac { 1 - \cos(2 \theta)} 2)^2 = \frac {1 - 2 cos(2 \theta) + \cos^2(2 \theta)} 4 = \frac {1 - 2 cos(2 \theta) + (\frac{1+\cos(4 \theta)}2)} 4 =\frac 3 8 - \frac {cos(2 \theta)} 4 + \frac {\cos(4 \theta)} 8$
• Dec 13th 2012, 03:22 AM
Tutu
Re: Integration, Trig substitution
Thank you! I got the first one!

I got $\displaystyle \frac{cot}{16}$ for the second one, and then after integrating and simplifying I got $\displaystyle \frac{1}{16}\ln\|\sin\theta|\+c$, could you show be how to further simplify?

For the third one, I could not get what you got.. for 4sin^2(delta)-2sin^4(delta) it seemed that you integrated to get that. But why? Later you performed another integration, so integrate two times?

Thank you really very much for your time!
J.
• Dec 13th 2012, 03:23 AM
Tutu
Re: Integration, Trig substitution
oops, it's (1/16) ln|(sin(delta)| + c, sorry!
• Dec 13th 2012, 05:32 AM
ebaines
Re: Integration, Trig substitution
For the second one, you need to replace the $\displaystyle \sin \theta$ term with the equivalent in terms of $\displaystyle x$:

$\displaystyle x = 4 \tan \theta = \frac {4 \sin \theta}{\cos \theta} = \frac {4 \sin \theta} {\sqrt {1 - \sin^2\theta}}$;

$\displaystyle x^2(1-\sin^2 \theta) = 16 sin^2 \theta$

Rearrange to get: $\displaystyle \sin \theta = \frac x {\sqrt {x^2+16}}$

So the result is

$\displaystyle \int \frac 1 {x (x^2 + 16)} dx = \frac 1 {16} ln(\frac x {\sqrt{x^2 + 16}}) + C$

For the third one - I guess I wasn't very clear. You said you reached a point where you need to integrate $\displaystyle 16 \cos^2 \theta - 16 \cos^4 \theta$ and you ended up with $\displaystyle 4 \sin (2 \theta) + 2 \sin (4 \theta) + C$. I arrived at a different expression to integrate, and then tried to show how to evaluate that integral. But as I review this I think your approach of integrating $\displaystyle 16 \cos^2 \theta - 16 \cos^4 \theta$ works as well. However, the result of that integral is not what you got - it should actually be: $\displaystyle 2 \theta - \frac 1 2 \sin(4 \theta)$.
• Dec 13th 2012, 08:58 AM
Tutu
Re: Integration, Trig substitution
Thank you so much!

I got the answer key for the questions, question 2 is right, thank you so much! It's just that it is |x| in the numerator but the denominator was $\displaystyle \sqrt{x^2+16}$, without the mod. I don't really see why..

For the third one, I cannot get what you got!
Here are my steps:

Simplifying $\displaystyle \16\cos^2\theta\-\16\cos^4\theta$
I have $\displaystyle \8\cos\2\theta\-\8\cos\4\theta$ left to integrate
After integration,

$\displaystyle \frac{8sin\2\theta}{2}-\frac{8sin\4\theta}{4} + c$
= $\displaystyle \4sin\2\theta-2sin\4\theta + c$
= $\displaystyle \8\sin\theta\cos\theta-4sin\2\theta\cos\2\theta + c$
that's where I'm stuck.
I think I had performed the integration wrongly..

Nonetheless, thank you so much!
• Dec 13th 2012, 09:27 AM
ebaines
Re: Integration, Trig substitution
Quote:

Originally Posted by Tutu
Simplifying $\displaystyle \16\cos^2\theta - 16\cos^4\theta$
I have $\displaystyle 8\cos2\theta-8\cos4\theta$ left to integrate

First, please note that some of you math symbols aren't working right, I believe because you are putting a backslash in front of minus signs and numbers where it is not needed.

I get $\displaystyle 16 \cos^2 \theta + 16 \cos^4 \theta = 2 - 2 \cos (4 \theta)$, which integrates easily enough to $\displaystyle 2 \theta - \frac {\sin (4 \theta)}2$