# Help with integration problem

• Sep 10th 2008, 07:21 AM
Matt164
Help with integration problem
This problem has been worked out for me, but I am confused about what exactly is going on in it.

Problem: Find the indefinite integral.
∫-1/sqrt(1-(2t-1)^2) dt

Step 1: Let u= 2t -1

Step 2: du= 2 dt

Step 3: ∫ -1/sqrt(1-(2t-1)^2) dt
= -1/2 ∫ 2/sqrt(1-(2t-1)^2) dt

Step 4: = -1/2 arcsin(2t-1) + C

The main thing I don't understand is where the
-1/2 before the ∫ sign is coming from in Step 3, and why the -1 changed to a 2 in the part after the ∫ sign. If anyone could explain to me what is going on in this problem I would greatly appreciate it. Thanks.
• Sep 10th 2008, 07:28 AM
Firstone
Lets say the denominator = x to save me from typing alot.

∫ -1/x
∫ -1/2 * 2/x

since -1/2 is a constant you may bring it to the front.
-1/2 ∫ 2/x

See how they are the same?
• Sep 10th 2008, 07:54 AM
Showcase_22
http://i116.photobucket.com/albums/o..._01/img029.jpg

Since dt/dx=(1/2)cosx, the 1/2 can be moved to the front of the integral as well as the -1. At the front of the integral this gives -1(1/2) which is -1/2.
• Sep 10th 2008, 08:16 AM
Chris L T521
Quote:

Originally Posted by Showcase_22
http://i116.photobucket.com/albums/o..._01/img029.jpg

Since dt/dx=(1/2)cosx, the 1/2 can be moved to the front of the integral as well as the -1. At the front of the integral this gives -1(1/2) which is -1/2.

Or you could let $u=2t-1\implies \,du=2\,dt$

The integral transforms into $-\tfrac{1}{2}\int\frac{\,du}{\sqrt{1-u^2}}$, which we know to be $-\tfrac{1}{2}\sin^{-1}u+C$

So the final result would be $-\tfrac{1}{2}\sin^{-1}(2t-1)+C$.

--Chris