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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top