How do I solve this integral using integration by parts?

  • Thread starter Thread starter peace-Econ
  • Start date Start date
  • Tags Tags
    Integral
peace-Econ
Messages
34
Reaction score
0

Homework Statement



Solve the integral.

Homework Equations



Integral: (x^m)*(1-x)^k where m is a nonnegative integer and k > -1

The Attempt at a Solution



I've tried to take this integral by using integral by parts, but I couldn't take it. Can anyone tell me how to take this integral? I really appreciate that.
 
Physics news on Phys.org
If it's an indefinite integral then that's an incomplete beta function. You can do a lot of things with it, but you can't write a simple elementary form for it. What do they really want you to do?
 
Sorry. the integral is actually 1 to 0. This question is actually induction.

Integral(1-0): (x^m)*(1-x)^k=n!/(k+1)(k+2)...(K+m+1) where m is a nonnegative integer and k > -1

So, I thought that if I take integral from the right side, I can prove it. But it does not seem the case...
 
peace-Econ said:
Sorry. the integral is actually 1 to 0. This question is actually induction.

Integral(1-0): (x^m)*(1-x)^k=n!/(k+1)(k+2)...(K+m+1) where m is a nonnegative integer and k > -1

So, I thought that if I take integral from the right side, I can prove it. But it does not seem the case...

Then start working on the integration by parts idea. Call your integral I(m,k). Work out m=0. For practice try doing small values of m (m=1, m=2, m=3) until you see what's going on. Then try to express I(m+1,k) in terms of I(m,k) and apply induction.
 
Last edited:
actually, I think I made it. Thanks!
 
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