Integration Q

[Sonal7]

I cant work out how they did the partial fraction decomposition. I also cant work out the integral for limits 1 and 0. I have I am getting ans in decimal but i suspect is a


whole number of a fraction they are looking for.

 

[topsquark]

They didn't use partial fractions, they just jumped a step.
[math]\dfrac{1}{t} \cdot \dfrac{1 - t^2}{1 - t^2 + 1 + t^2} = \dfrac{1}{t} \cdot \dfrac{1 - t^2}{2}[/math]
[math]= \dfrac{1- t^2}{2t}[/math]
Can you finish?

-Dan

 

[Sonal7]

I will try thank you. I got it now as you just split up the terms on the top and then cancel the t in one of the new terms.

I now need a trick for the integral. I think there is.

 

[topsquark]

I will try thank you. I got it now as you just split up the terms on the top and then cancel the t in one of the new terms.

I now need a trick for the integral. I think there is.

What trick do you think you need? It's a couple of basic integrals:
[math]\int \dfrac{1 - t^2}{2t} ~ dt = \int \dfrac{1}{2t} ~ dt - \int \dfrac{t}{2} ~ dt[/math]
[math]\int \dfrac{1}{t} ~ dt = ln|t| + C[/math]
[math]\int t ~ dt = \dfrac{1}{2} t^2 + C[/math]
-Dan

 

[Sonal7]

Yes I got that. I don't understand how you get the integral as I am getting ans in decimals if you use the calculator but I think its a whole number. The calculator is not smart you see. I have an idea I will to sub in tan 0.5 and tan 0.25 but I cant figure that out.

 

[Sonal7]

Thank you very much Dan, atleast I got the main part right. I am just thinking of what they are getting at with the limits of tan .5 and tan 0.25.I slept over it and its not come to me yet.

 

[topsquark]

Thank you very much Dan, atleast I got the main part right. I am just thinking of what they are getting at with the limits of tan .5 and tan 0.25.I slept over it and its not come to me yet.

[math]t = tan \left ( \dfrac{1}{2} x \right )[/math]
So, for example, if we have x = 1 then t = tan(1/2) = tan(0.5), so the upper limit on the t integration would be tan(0.5).

-Dan

 

[Sonal7]

Yes i was wrong, the ans is in decimals.