Confused on what they substitued answer given.

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In summary: 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? Yes, it would.
  • #1
mr_coffee
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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 [Broken]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
 
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  • #2
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.
 
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  • #3
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;
 
  • #4
mr_coffee said:
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
 
  • #5
mr_coffee said:
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
 
  • #6
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
 

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When someone says they are "confused on what they substituted answer given," it means that they are unsure or uncertain about what has been used to replace the original answer or solution that was given.

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