# Thread: evaluating integrals - quick question help

1. ## evaluating integrals - quick question help

evaluate integrals:

{ (pie/2,0) ( (1+cos2t)/2 )dt

cos2t = 2 (cost)^2 - 1

so,

{ (pie/2,0) (cost)^2 dt

but how do you approach this problem after this step?

i forgot the anti derivative of (cost)^2 ...someone help ~ do it for me so i can see

thx

2. $\displaystyle \cos2t=\cos^2t-\sin^2t=2\cos^2t-1.$

As for your last question, put $\displaystyle f(x)=\int_{0}^{\sqrt{x}}{\cos t\,dt}=\sin \sqrt{x}$ thus $\displaystyle f'(x)=\frac{\cos \left( \sqrt{x} \right)}{2\sqrt{x}}.$ Hence, by using the FTC we have $\displaystyle f'(x)=\cos \left( \sqrt{x} \right)\cdot \left( \sqrt{x} \right)',$ which leads the same result.

3. bump *new question

4. You don't need to integrate that, if you can prove that $\displaystyle \int_{0}^{\frac{\pi }{2}}{\cos ^{2}t\,dt}=\int_{0}^{\frac{\pi }{2}}{\sin ^{2}t\,dt}$ then the result follows easily.

5. what do you mean? i don't get it...

it says evaluate the integral

how do i do it !?

someone please do this one before i have to go !

ty

6. Originally Posted by pghaffari
what do you mean? i don't get it...

it says evaluate the integral

how do i do it !?

someone please do this one before i have to go !

ty
By editing your original question multiple times the replies that you get make no sense to someone tying to read and follow the thread. This is annoying and insulting to the members of MHF.

Do not edit your question once you get replies. If you have new questions, ask them in a new thread. If you have follow-up questions, ask them in the same thread - do NOT edit your original question to ask them.