Expert: Ahmed Salami - 12/5/2004

Question
I cannot seem to find the integral of Sin^-1 x by dx when the boundaries are 0 and 1.

Also, I could not figure out the boundaries of the "loop" in the function y^2 = x^5 + 4x^4.
Can you help? Thanks!!

Answer
Hi Jennifer,
Really sorry for the delay, i am having my exams and doing a project study at the moment.
To integrate arcsin x dx, we use the substitution
u = arcsin x
x = sin u
dx = cos u du
Therefore,
{arcsin x dx = {u.(cos u) du
            = u(sin u) + cos u
but cos u = sqrt(1 - sin^2 u) = sqrt(1 - x^2)
{arcsin x dx = (arcsin x). x + sqrt(1 - x^2)
Inserting the limit 1 gives #/2 . 1 + 0
= #/2    where # = pi
inserting 0 gives 0 + 1 = 1
The final answer is thus,
#/2 - 1

As for the other question, differentiating gives
2y dy/dx = 5x^4 + 16x^3
dy/dx = (5x^4 + 16x^3)/ 2sqrt(x^5 + 4x^4)
The max. and min. values would occur when dy/dx = 0
i.e at 5x^4 + 16x^3 = 0
which gives x = 0 and -16/5
the right and left boundaries occur when dy/dx is infinite
i.e at 2sqrt(x^5 + 4x^4) = 0
or simply x^5 + 4x^4 = 0
which gives x = 0 and -4
I think that should help.
I hope you get it. You can always get back to me.
Regards.