How do I integrate a difficult integral with integration by parts?

[sara_87]

Homework Statement

I need to evaluate the integral:


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

Homework Equations

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

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.

 

[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.

 

[Mark44]

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

 

[blake knight]

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

 

[Mark44]

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

Good tip.

 

[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.

 

[sara_87]

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

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

 

[blake knight]

This is going to be a part of the integration by parts.

 

[blake knight]

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

This is going to be a part of integration by parts.

 

[sara_87]

so then u would be:
\frac{1}{(a^2-x^2)^{1-c}}
??
i don't think, since in the numerator, we have: (x^-c)' what happens to the derivative?
i think I am confused

 

[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

 

[blake knight]

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

 

[sara_87]

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

 

[sara_87]

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

what then do i do with the derivative part?

 

[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.

 

[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?

 

[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).

 

[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
instead, i am getting:
2(c-1)(a)^(b-2)(1-sin(y))^(c-2)cos(y) dy
??

 

[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.

 

[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 I am trying to solve this integral...and its a pain.

 

[blake knight]

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 I am trying to solve this integral...and its a pain.

Ok so then you'll agree that if v'=(x^-c)' then v=(x^1-c)/1-c<----which is what I suggested before...Only that after Integrating by parts youll end up having an Integrand susceptible for Integration by parts again, and to infinity. This case happens due to the fact that we have a variable as an exponent. The equation can be solve by setting c not equal to zero and setting a and b to infinite.

 

[sara_87]

if v'=(x^-c)' then v=(x^-c)
but i am now using the method suggested by MathematicalPhysicist (the first reply) and finally i am getting: cos^(2(c-2)+1)(y)/sin^(c-1)(y) dy but I am still doing the integration by parts to try and get the integrand (cosy)/((siny)^(c-n))

but how would this lead to the hypergeometric 2F1

 

[sara_87]

repeat this process until you get an integral with an integrand cos(y)/sin^(c-n)(y) which is easy to compute.

I am not getting cos(y) on its own, it's always to some power in terms of c.