# Trig substitution

1. Mar 12, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
consider the region bounded by the graphs of y=arcsinx, y=0, x = 1/2.

a) find the area of the region.
b) find the centroid of the region.

2. Relevant equations

$\displaystyle\int_0^{1/2} {arcsinx dx}$

u=arcsinx; du = $\frac{1}{1-x^2}dx$

dv=dx ; v=x

xarcsinx]$^{1/2}_{0}$ - $\displaystyle\int_0^{1/2} {\frac{x}{\sqrt{1=x^2}} dx}$

= (1/2arcsin(1/2) - 0) + $\sqrt{1-x^2}^{1/2}_{0}$

answer: $\frac{\pi + 6\sqrt{3}-12}{12}$

3. The attempt at a solution

in my work i have:

xarcsix]$^{1/2}_{0}$ = $\frac{\pi}{12}$

for the other part i get:

$\sqrt{1-x^2}^{1/2}_{0}$ = $\sqrt{1-\frac{1}{4}} - 1$

which i can see how the last term -1 will go to -12/12, and pi/12 goes if front, but how are the getting the middle term, 6$\sqrt{3}$?

Last edited: Mar 12, 2013
2. Mar 12, 2013

### SammyS

Staff Emeritus
It seems to me that this would be much easier to do by looking at x as a function of y.

3. Mar 12, 2013

### iRaid

I think for this problem, it's just easier to do a substitution at the beginning, x=sinθ. Then you just do a integration by parts.

Note this only works because the limits of integration are between 0 and 1/2. Also, arcsin(sinx)=x.

4. Mar 12, 2013

### eumyang

The middle "term" you want to get is $\frac{6\sqrt{3}}{12}$, not $6\sqrt{3}$.

From $\sqrt{1-\frac{1}{4}}$, do the subtraction, and split into two square roots, one in the numerator, and one in the denominator. It's straightforward from there.

5. Mar 12, 2013

### whatlifeforme

solved.thanks.