Confused on what they substitued answer given.

[mr_coffee]

Hello everyone I'm studying for my calc 3 exam and I'm lost how they went from this substitution, my professor did the work.

Here is the problem and the answer under it:
http://img150.imageshack.us/img150/606/sssk5.jpg Well he showed the following:1/2* integral 0 to PI/2; integral 0 to 1 u*cos(uv) dv du;

let t = uv;

then he got:

1/2 integral 0 to PI/2 u * 1/u sin(uv); now plug in 0 and 1 and he got:

1/2 * integral 0 to PI/2 sin(u) du = 1/2 cos(u) and plug in PI/2 and 0 and u get 1/2.

But what I'm lost on is, if you let t = uv;

then if you took the derivative of t, with respect to v you would get 0 wouldn't you? or do i take the partial deriviatve? if i took the partial derivative with respect to v, then i would get t = u; which would make sense i think...maybe not

 

[JasonRox]

You have x-y=u and x+y=v.

Ok, I'm so lost at how you wrote this stuff.

Try writing integrals as bIa for integrating from b to a, for future reference. And show you du's and dv's because I'll get lost without those going around and other people would get lost too probably.

 

[mr_coffee]

right but then when you sub u in for x-y and v in for x + y your left with:
u*cos(uv) dv du;

and you have to make another substiution to do that integral, t = uv; which is where I get lost:
t = (x-y)(x+y) ?
t = x^2-y^2;

 

[d_leet]

right but then when you sub u in for x-y and v in for x + y your left with:
u*cos(uv) dv du;

and you have to make another substiution to do that integral, t = uv; which is where I get lost:
t = (x-y)(x+y) ?
t = x^2-y^2;

You don't absolutely need a substitution to do that integral but if you can see it better by making one then if you have t=uv, then dt=udv. And you have the integral of cos(t)dt

 

[JasonRox]

if you let t = uv;

then if you took the derivative of t, with respect to v you would get 0 wouldn't you?

No, it's not zero.

dt/dv = u

 

[mr_coffee]

Thanks guys,

now i have

t = uv;
dt/dv = u dv;

cos(t) integral of that is sin(t), sub back in t = uv; and u get:
sin(uv) now plug in 1 and 0 for v and your left with:
integral 0 to PI/2 sin(u) du; but isn't the integral of sin(u) = -cos(u)?

so wouldn't it be -1/2 as the naswer not 1/2?

I understand u*1/u = 1; But when I did it, I didn't use that step at all to get sin(uv) after the subbing

He had a step which I didn't quite understand..he said:

1/2 integral 0 to PI/2 u*1/u sin(uv) du