Integration by parts & inv. trig fxn

wvcaudill2
Messages
53
Reaction score
0

Homework Statement


\int xarcsin2xdx

2. The attempt at a solution
Image1.jpg


Can someone explain to me what is happening at step 2? I understand how the integration by parts was done, but where does the (1/8) or (2x) come from?
 
Physics news on Phys.org
They multiplied the numerator and denominator by 8. Thus

\frac{x^2}{\sqrt{1-4x^2}}=\frac{8x^2}{8\sqrt{1-4x^2}}

and then they rewrote this slightly.
 
micromass said:
They multiplied the numerator and denominator by 8. Thus

\frac{x^2}{\sqrt{1-4x^2}}=\frac{8x^2}{8\sqrt{1-4x^2}}

and then they rewrote this slightly.

I still don't see how this gets (1/8), and even then, what is the purpose of doing this, and where does the (2x) come from?
 
Yoi get the 1/8 by bringing the denominator out of the integral.
And the numerator is rewritten as 8x^2=2(2x)^2.

The purpose of doing that is to do a substitution t=2x.
 
micromass said:
Yoi get the 1/8 by bringing the denominator out of the integral.
And the numerator is rewritten as 8x^2=2(2x)^2.

The purpose of doing that is to do a substitution t=2x.

Oh, ok. Thanks again!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top