Evaluating Integrals with U-Substitution

[SeekerOfDragons]

I'm having issues solving the following problem:

Integrate ( (Pi^2)/36 , (Pi^2)/4 ) [ Cos( sqrt(t) ) / sqrt( t * Sin( sqrt(t) ) ] dt

I'm thinking I need to set u = t * Sin( sqrt(t) )

If that is true, then I'm having issues solving for du

du = t * Cos( sqrt(t) ) * 1/2 t^(-1/2) + Sin( sqrt(t) ) dt

du = [ t * Cos (sqrt(t) ) / (2 sqrt(t) ) ] + Sin( sqrt(t) ) dt

du = [ t * Cos (sqrt(t) ) / (2 sqrt(t) ) ] + ( 2 sqrt(t) * Sin( sqrt(t) ) / 2 sqrt(t) ) ] dt

du = ( 2 * sqrt(t) * Sin( sqrt(t) ) + t * Cos( sqrt(t) ) / 2 sqrt(t) ) dt

2 sqrt(t) du / (t * Cos( sqrt(t) ) + 2 sqrt(t) * Sin( sqrt(t) ) = dt

I'm not sure if I did the derivative correct or if I chose the wrong thing to substitute for u because what I came up with for dt above won't really help cancel anything out in the above equation. if I did my math wrong and it's really supposed to be du/cos( sqrt(t) ) = dt then maybe I can have an easier time at it.

any pointers or advice on where I went wrong and where I need to go would be greatly appreciated on this one.

 

[galactus]

Sub \(\displaystyle u=sin(\sqrt{t}), \;\ 2du=\frac{cos(\sqrt{t})}{\sqrt{t}}dt\)

and it falls into place.

 

[SeekerOfDragons]

That works. The hardest part I have with U-Substitution is figuring out what to set U equal to.

Integrate ( (pi^2)/36, (pi^2)/4) Cos( sqrt(t) ) / sqrt( t * Sin( sqrt(t) ) ) dt
Integrate ( (pi^2)/36, (pi^2)/4) Cos( sqrt(t) ) / sqrt(t) * sqrt(Sin( sqrt(t) ) dt

setting U = Sin( sqrt(t) ) :

du = Cos( sqrt(t) ) * (1/2) t^(-1/2) dt
du = Cos( sqrt(t) ) / 2 sqrt(t) dt
2du = Cos( sqrt(t) ) / sqrt(t) dt

Integrate (?, ?) [ 2du / sqrt(u) ] du
2 Integrate (?, ?) [ U^(-1/2) ] du

2 * ( U^(1/2) / (1/2) )
4 U^(1/2)

4 Sin( sqrt(t) ) ^ (1/2) ] (pi^2/36, pi^2/4)

leaving the correct answer to be: 4 * Sin( sqrt(pi^2/4) ) - 4 * Sin( sqrt(pi^2/36) )

did i compute all that correctly?

 

[BigGlenntheHeavy]

\(\displaystyle \int_{\pi^{2}/ 36}^{\pi^{2}/ 4}\frac{cos(t^{1/2})}{t^{1/2}(sin(t^{1/2}))^{1/2}}dt\)

\(\displaystyle Let \ u \ = \ sin(t^{1/2}), \ then \ du \ = \ \frac{cos(t^{1/2})}{2t^{1/2}}dt\)

\(\displaystyle 2t^{1/2} du \ = \ cos(t^{1/2})dt\)

\(\displaystyle Hence, \ we \ have \ 2\int_{1/2}^{1}\frac{t^{1/2}du}{t^{1/2}u^{1/2}} \ = \ 2\int_{1/2}^{1}u^{-1/2}du\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

 

[SeekerOfDragons]

Big Glenn,

I had managed to get to 2 * Integral (a, b) u^(-1/2) du which became

4 u^(1/2) ] (a, b)

I had elected to sub back u as Sin(sqrt(t)) and evaluate it with the original points, unfortunately when I input the figures into my calculator, I had an input error on my part (namely not taking the square root of Sin( sqrt() ) which threw off my answer completely) and wasn't getting a correct answer. going back and doing it the way you suggested with a=1 and b=1/2 and then with a=Pi^2/4 and b=Pi^2/36 and putting the formula correctly into my calculator, I came up with a final answer of:

4 - 2 * sqrt(2)