# Trigonometric substitution, integral of 8dx/(4x^2+1)^2 basic error somewhere

• Feb 18th 2011, 08:00 PM
Wolvenmoon
Trigonometric substitution, integral of 8dx/(4x^2+1)^2 basic error somewhere
Alright...been at this on and off all day!

The problem is the integral of 8dx/(4x^2+1)^2.

I let x=1/2*tanT

dx=sec^2(T)/2

I work this problem, going through the integral of 4sec^2T/sec^4T

Push through to integral of 4*cos^2T, skipping forward I end up with 2t+sin(2t)

I build my triangle. Side opposite is 2x. Side adjacent is 1. Hypotenuse is 4x^2+1.

T=arctan(2x). Sin(2t) = side opposite (2x) over hypotenuse ( 4x^2+1 ).

So my answer is 2arctan(2x)+2x/(4x^2+1)+C However I have a problem. In the second term there, that should be 4x.

Any idea where I missed it?

Edit:

Alright, second problem. Integrating x^2/(1-x^2)^(5/2)dx

Substitutions are x=sint
dx=cost(t)

I work it all the way down to the integration of tan^2tsec^2t dt=tan^3t/3.

My triangle is side opposite=x, side adjacent= (1-x^2), hypotenuse 1. I substitute in and get (1/3)*(x/(1-x^2))^3 + C

Once again my side adjacent is messed up. The side adjacent in my answer should be the square root of what I have.

I think the problems are related.
• Feb 18th 2011, 08:20 PM
Prove It
OK, for starters, you're mixing up your x's and t's...

$\displaystyle \int{\frac{8\,dx}{(4x^2 + 1)^2}}$

Make the substitution $\displaystyle x = \frac{1}{2}\tan{t} \implies dx = \frac{1}{2}\sec^2{t}\,dt$ and the integral becomes

$\displaystyle \int{\frac{8\cdot \frac{1}{2}\sec^2{t}\,dt}{\left[4\left(\frac{1}{2}\tan{t}\right)^2 + 1\right]^2}}$

$\displaystyle = \int{\frac{4\sec^2{t}\,dt}{(\tan^2{t} + 1)^2}}$

$\displaystyle = \int{\frac{4\sec^2{t}\,dt}{\sec^4{t}}}$

$\displaystyle = \int{\frac{4\,dt}{\sec^2{t}}}$

$\displaystyle = \int{4\cos^2{t}\,dt}$

$\displaystyle = \int{2 + 2\cos{2t}\,dt}$

$\displaystyle = 2t + \sin{2t} + C$

$\displaystyle = 2t + \frac{2\tan{t}}{1 + \tan^2{t}} + C$

$\displaystyle = 2\arctan{2x} + \frac{4x}{1 + (2x)^2} + C$

$\displaystyle = 2\arctan{2x} + \frac{4x}{1 + 4x^2} + C$.
• Feb 18th 2011, 08:24 PM
Wolvenmoon
3rd line from the bottom and the second term. What's happening there?

I also added a second problem to the first post.

I'll go fix my variables. I've been going a bit too long today. Thanks for your help!
• Feb 18th 2011, 08:48 PM
Prove It
It's a standard trigonometric identity: $\displaystyle \sin{2x} \equiv \frac{2\tan{x}}{1 + \tan^2{x}}$
• Feb 18th 2011, 10:05 PM
Wolvenmoon
Man, you're fast. Any help on the second one? ( I had to stop for the night due to neck and back inflammation )
• Feb 18th 2011, 10:16 PM
Prove It
You really should make a new thread for new questions... But anyway...

$\displaystyle \int{\frac{x^2\,dx}{(1 - x^2)^{\frac{5}{2}}}}$.

Make the substitution $\displaystyle x = \sin{t} \implies dx = \cos{t}\,dt$ and the integral becomes

$\displaystyle \int{\frac{\sin^2{t}\cos{t}\,dt}{(1 - \sin^2{t})^{\frac{5}{2}}}}$

$\displaystyle = \int{\frac{\sin^2{t}\cos{t}\,dt}{(\cos^2{t})^{\fra c{5}{2}}}}$

$\displaystyle = \int{\frac{\sin^2{t}\cos{t}\,dt}{\cos^5{t}}}$

$\displaystyle = \int{\frac{\sin^2{t}\,dt}{\cos^4{t}}}$

$\displaystyle = \int{\tan^2{t}\sec^2{t}\,dt}$

Now make the substitution $\displaystyle u = \tan{t} \implies du = \sec^2{t}\,dt$ and the integral becomes

$\displaystyle \int{u^2\,du}$

$\displaystyle = \frac{1}{3}u^3 + C$

$\displaystyle = \frac{1}{3}\tan^3{t} + C$

$\displaystyle = \frac{1}{3}\left(\pm \frac{\sin{t}}{\sqrt{1 - \sin^2{t}}}\right)^3 + C$

$\displaystyle = \frac{1}{3}\left(\pm \frac{x}{\sqrt{1 - x^2}}\right)^3 + C$

$\displaystyle = C \pm \frac{x^3}{3\sqrt{(1-x^2)^3}}$.
• Feb 18th 2011, 10:32 PM
Wolvenmoon
Thanks! I'm afraid of cluttering up the forums.

So, what's happening to the tangent between:

\displaystyle = \frac{1}{3}\tan^3{t} + C

and

\displaystyle = \frac{1}{3}\left(\pm \frac{\sin{t}}{\sqrt{1 - \sin^2{t}}}\right)^3 + C ?

What rule is being used there?

P.S., what's the easiest way to write Latex for this forum?
• Feb 18th 2011, 10:35 PM
Prove It
Another standard identity:

$\displaystyle \tan{x} \equiv \frac{\sin{x}}{\cos{x}} \equiv \frac{\sin{x}}{\pm \sqrt{1 - \sin^2{x}}}$...
• Feb 18th 2011, 10:47 PM
Wolvenmoon
Okay, I knew that one just not that I'd be applying it in that way. My instructor sent us in a bit blind here. I think it might have been unintentional.
• Feb 18th 2011, 10:57 PM
Prove It
Quote:

Originally Posted by Wolvenmoon
Okay, I knew that one just not that I'd be applying it in that way. My instructor sent us in a bit blind here. I think it might have been unintentional.

I always just use trigonometric identites to write whatever I find the integral to be in terms of whatever I had originally substituted, so that I can back substitute.

List of trigonometric identities - Wikipedia, the free encyclopedia

• Feb 19th 2011, 03:25 PM
Wolvenmoon
Alright, on the second problem. I'm supposed to be using a triangle to substitute back.

So I take x=sint, this means that sint=x/1. So the side opposite on my triangle becomes X and my hypotenuse becomes 1.

Then I come back to the original problem and I take part of the denominator. From what he told us I got the gist that we ignored the square root at times when substituting back but not at other times.

What's the rule here? My formula sheet talks about integrands of (a^2-x^2)^(1/2) using a substitution of x=asint, one that's the root of a^2+x^2 that's x=atant, and one of the square root of x^2-a^2 and x=secT

When do I keep the root and when do I throw it out?