one small issue with trig substitution. need clarification.

[Necro]

I simply need an explanation, please dont simply resort to finishing the problem unless you want to be detailed in what the thought process is.

Anyway the problem:
integrate (100-x^2)^(1/2) dx rom 5 to 10(top number)

eventually i get the integral to simplify to: 100 integrate cos(theta)^2
(I will not do anything with the 100 going forward because my confusion is rooted elsewhere)

the integral of cosine theta is: (x/2) + sin(2x)/4 +c

sin(2x)/4 becomes (1/4)2sin(theta)cosine(theta) = (1/2)sin(theta)cosine(theta)

so im left with (1/2)x + (1/2)sin(theta)cosine(theta)

I dont know what to do with x/2. with sin and cosine i just goto my reference triangle. Im pretty sure thats what I do with x but i have no idea.

 

[studentMCCS]

It should be theta/2, not x/2.

u=theta

I think you have to go back to where you said x=10sin(u), solve for theta, and then substitute.

x/10=sin(u)
u=Asin(x/10)

Now your going from 5-10.

So first you will plug in 10 for x.

arcsin(1)=?

sin(what)=1

look at unit circle an you see that at pi/2, sin=1.

and arcsin(1/2)=?
sin(what)=(1/2)
look at unit circle, and you see it will be at pi/6 where sin is 1/2.

I think this is right.

 

[Necro]

It should be theta/2, not x/2.

u=theta

I think you have to go back to where you said x=10sin(u), solve for theta, and then substitute.

x/10=sin(u)
u=Asin(x/10)

so then:


(1/2)theta + (1/2)sin(theta)cosine(theta)


becomes


(1/2)asin(x/10) + (1/2)(x/10)(100-x^2)^(1/2)
not simplified so that logic is more clear

correct?