Substitution with trigonometric identities - really confused!

[wduk]

Hi

I've been reading about doing substitutions for integration, but it just isn't sinking in my brain properly on how to find the right option to substitute. I am not understanding from my book how the examples are picking what they pick.

So for example there is the integral of:

1 / ( x^2 * sqrt(1-4x^2) dx

The example has:

Let x = 1/2 * sin(u), so dx/du = 1/2*cos(u) there for u = sin^-1(2x).

I understand that a^2 - x^2 = a*sin(u) for substitution. But am not understanding where the 1/2 coefficient came from for x.

Wouldn't 1-4x^2 = 1^2-(2x)^2 = 1*2sin(u) not 1/2sin(u) ?

 

[HallsofIvy]

Hi

I've been reading about doing substitutions for integration, but it just isn't sinking in my brain properly on how to find the right option to substitute. I am not understanding from my book how the examples are picking what they pick.

So for example there is the integral of:

1 / ( x^2 * sqrt(1-4x^2) dx

The example has:

Let x = 1/2 * sin(u), so dx/du = 1/2*cos(u) there for u = sin^-1(2x).

I understand that a^2 - x^2 = a*sin(u) for substitution. But am not understanding where the 1/2 coefficient came from for x.

Wouldn't 1-4x^2 = 1^2-(2x)^2 = 1*2sin(u) not 1/2sin(u) ?

No, if x= 2sin(u) then \(\displaystyle x^2= 4 sin^2(u)\) so that \(\displaystyle 1- 4x^2= 1- 4(4sin^2(u))= 1- 16sin^2(u)\).

With x= (1/2)sin(u), \(\displaystyle x^2= (1/4)sin^2(u)\) and \(\displaystyle 1- 4x^2= 1- 4((1/4)sin^2(u))= 1- sin^2(u)= cos^2(u)\).

 

[wduk]

Ahh that makes more sense now. Gosh i am really bad at working out the right substitution for some reason with these, i was doing fine with simpler ones