Thread: tricky (?) intergration with inclusion of arcsin(x-3)

i am getting myself a bit lost with the integration of a function P(x), i have tried to solve it by integration by parts, but then end up needing to integrate arcsin(x-3), which is proving trickier than i was expecting, i have come up with an answer but its not give the results that i expect, i'e between the limits of 2 and 4 the answer should be 3. would anyone mind looking at the attached PDF and give me a few pointers where i am going wrong
Cheers
WN

Hi

Let u = 3-x

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u-3}{\sqrt{1-u^2}}\:du$

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u}{\sqrt{1-u^2}}\:du - 3 \int \frac{1}{\sqrt{1-u^2}}\:du = -\sqrt{1-u^2} - 3 Arcsin u$

Originally Posted by running-gag

Hi

Let u = 3-x

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u-3}{\sqrt{1-u^2}}\:du$

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u}{\sqrt{1-u^2}}\:du - 3 \int \frac{1}{\sqrt{1-u^2}}\:du = -\sqrt{1-u^2} - 3 Arcsin u$

sorry you have lost me, where does this come in to it, is this the whole answer, the answer after the second intergration by parts, or where i am doing the substitution at the end. i can't seem to find where is it.
also you have u=3-x, then in the second equation you have x=u-3 which isn't right as from the definition on u, x-3-u
Hope you can clarify that
Cheers
WN

I only used substitution (no integration by parts)

u = 3-x gives x=3-u and dx=-du therefore x dx = (u-3) du

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u-3}{\sqrt{1-u^2}}\:du$

Then I split the integral in 2 parts

$\displaystyle \int \frac{x}{\sqrt{1-(3-x)^2}}\:dx = \int \frac{u}{\sqrt{1-u^2}}\:du - 3 \int \frac{1}{\sqrt{1-u^2}}\:du = -\sqrt{1-u^2} - 3 Arcsin u$

Originally Posted by running-gag

I only used substitution (no integration by parts)

ah yes i see now, thanks a lot.
WN