# Difficult integration

1. Oct 14, 2009

### sara_87

1. The problem statement, all variables and given/known data

I need to evaluate the integral:

$$\int^a_b \frac{(x^{-c})'}{(a^2-x^2)^(1-c)} dx$$

2. Relevant equations

(x-c)' means the derivative with respect to x.

3. The attempt at a solution

I tried to use integration by parts:

let u=(a2-x2)c-1 so u'=(c-1)(a2-x2)c-2(-2x)
and v'=(x-c)' so v=x-c

the integral is then:

$$\left[uv\right]^a_b+\int^a_b\frac{2(c-1)(a^2-x^2)^{c-2}}{x^{c-1}}dx$$

I can evaluate the uv part but i am struggling to integrate
$$\int^a_b\frac{2(c-1)(a^2-x^2)^{c-2}}{x^{c-1}}dx$$
any help would be very much appreciated. thank you.

2. Oct 14, 2009

### MathematicalPhysicist

Why not try trig substitution, x/a=sin(y), you'll get something of the sort of:
cos^(2(c-2)+1)(y)/sin^(c-1)(y) dy (with some factors), now repeat integration by parts by v'=cos(y)/sin^(c-1)(y) u=cos^(2(c-1))(y), repeat this process until you get an integral with an integrand cos(y)/sin^(c-n)(y) which is easy to compute.

3. Oct 14, 2009

### Staff: Mentor

At first I though you could simplify the problem by bringing the (1 - c) factor out of the integrand, but then I discovered that (1 - c) is an exponent, not a factor.
Your LaTeX was very well done except for that part. For exponents in LaTeX expressions, surround them with braces -{}- instead of parentheses, in all cases where the exponent is more than a single character.

Here's the revised integral.
$$\int^a_b \frac{(x^{-c})'}{(a^2-x^2)^{1-c}} dx$$

4. Oct 14, 2009

### blake knight

One thing: If you set v'=x^-c; then v=x^1-c/(1-c)

5. Oct 14, 2009

### Staff: Mentor

Good tip.

6. Oct 14, 2009

### sara_87

thanks all, i will try now to integrate it using your advice. mark44, you're right, i did make a mistake in Latex, it should be ^(1-c) thanks.

7. Oct 14, 2009

### sara_87

thanks, so, would this be substitution or is this part of the integration by parts?

8. Oct 14, 2009

### blake knight

This is gonna be a part of the integration by parts.

9. Oct 14, 2009

### blake knight

This is gonna be a part of integration by parts.

10. Oct 14, 2009

### sara_87

so then u would be:
$$\frac{1}{(a^2-x^2)^{1-c}}$$
??
i dont think, since in the numerator, we have: (x-c)' what happens to the derivative?
i think im confused

11. Oct 14, 2009

### blake knight

Remember this? : 3. The attempt at a solution

I tried to use integration by parts:

let u=(a2-x2)c-1 so u'=(c-1)(a2-x2)c-2(-2x)
and v'=(x-c)' so v=x-c

12. Oct 14, 2009

### blake knight

I just replace your v=x^-c by v=x^1-c/1-c.

13. Oct 14, 2009

### sara_87

yes but you said that i should use v'=x^(-c) not v'=[x^(-c)]'
?

14. Oct 14, 2009

### sara_87

what then do i do with the derivative part?

15. Oct 14, 2009

### blake knight

You will arrive at having: = [(x^1-c/1-c)*(a^2-x^2)^c-1] - [integral of (x^1-c/1-c)*(-2x)(c-1)(a^2-x^2)^c-2 dx]--->continue from here.

16. Oct 14, 2009

### sara_87

but when we have an integral, and we want to use integration by parts, say:
$$\int f(x)g(x)$$
then u=f(x) and v'=g(x)

here my f(x)=(x-c)'
and g(x)=(a2-x2)c-1

so we said that u=(a2-x2)c-1
and so our v' must be (x-c)'
so why are we taking v'=(x-c)
am i wrong?

17. Oct 14, 2009

### blake knight

Oh I see what you are getting now : I thought there is a small 1 there in v'=(x^-c)^1 which of course is just equal to v'=(x^-c).

18. Oct 14, 2009

### sara_87

oh i see what you mean too. no that's not a 1 it's a derivative, that's what i meant when i said 'what do i do with the derivative'.
:)
but i tried to do it using the advice from MathematicalPhysicist (the first reply)
but i am not getting
cos^(2(c-2)+1)(y)/sin^(c-1)(y) dy (with some factors)
as he/she mentioned
2(c-1)(a)^(b-2)(1-sin(y))^(c-2)cos(y) dy
??

19. Oct 14, 2009

### blake knight

Im a little bit confused too, I know we are going too far now...but its a little bit strange to me when you have a derivative sign for your v', like v'=(x^-c)' and have a dx at the end of the integral, Isnt it repeating? I've never seen an equation of this form before. I maybe wrong.

20. Oct 14, 2009

### sara_87

i put v'=(x^-c)' just to get rid of the derivative but not getting very far.
it is a difficult integral, the solution contains the hypergeometric 2F1
but im trying to solve this integral...and its a pain.