1. ## intergration

im stuck on a intergration question as usual.

Evaluate this function

$K = \int\limits_{0}^{\frac{pi}{4}} sin2t dt$

any help would be seriously appreciated guys thanks

2. $y' = \sin (2x)\quad \Rightarrow \quad y = \frac{{ - \cos (2x)}}{2} + C$

3. I assume you know the INT.[sinU]dU.
It is -cosU +C.

Now, in your problem here, if U=2t, what would be the dU?
It is 2dt.

But you have only dt or 1dt in the integrand.
If you make the dt as 2dt, then you have INT.[sin(2t)]2dt. Which you can integrate easily. The answer should be -cos(2t) +C.

But you cannot make the dt as 2dt just like that. That is not allowed, or the new integrand is not the same as the original anymore.

So to maintain the value of the original integrand, you should have multiplied the original dt by (2/2), which is equal to 1 anyway.
Like so:
= INT.(0-->pi/4)[sin(2t)](2/2)dt
= INT.(0-->pi/4)[sin(2t)](2dt)/2
= (1/2)INT.(0-->pi/4)[sin(2t)](2dt)
= (1/2)[-cos(2t)] | (0-->pi/4)
= -(1/2)[cos(2pi/4) -cos(2*0)]
= -(1/2)[cos(pi/2) -cos(0)]
= -(1/2)[0 -1]