# 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