What is F'(x) when F(x)=Integral from 0 to x of (sin^2(t) dt)

Question

anonymous · Answer

$$F(x)=\int\limits_{0}^{x}\sin^2(t)dt$$
What is F'(x)

anonymous · Answer

you need to use a trig identity for sin^2(t)

anonymous · Answer

what is a trig identity?

anonymous · Answer

you are in calculus but you do not know what a trig identity is?

anonymous · Answer

im bad with terminology. I thought that the derivative of sin^2(t) was just 2cost

anonymous · Answer

$$\sin^{2}(x) = \frac{1}{2}(1 + \cos(2t))$$

anonymous · Answer

you can replace sin with the right side of the equality

anonymous · Answer

ok but what about the fact that it's F(x) and it's t in the function?

anonymous · Answer

just replace the x with a t

anonymous · Answer

oh lol ok

anonymous · Answer

i didn't know that thx

anonymous · Answer

$$F(x)=\int\limits\limits_{0}^{x}\frac{1}{2}(1 + \cos(2t))dt$$

anonymous · Answer

$$F(x)=\frac{1}{2}\int\limits\limits\limits_{0}^{x}(1 + \cos(2t))dt$$

anonymous · Answer

so $$F'(x)=\int\limits_{0}^{x}2\cos(2t)$$ Right?

anonymous · Answer

no you need to integrate the first function

anonymous · Answer

so $$F(x)=1/2(t-\sin(2t)/2)$$

anonymous · Answer

yeah no plug in the upper and lower limits

anonymous · Answer

yeah i guess my identiy was a little of but it looks like you corrected it

anonymous · Answer

ok so then now i can find f'(x) right?

anonymous · Answer

which would be 1/2-cos(2t)/2 right?

anonymous · Answer

yes but you were supposed to plulg in the upper and lower limits which would of changed it to x

anonymous · Answer

ok, so it would be$$F'(x)=(1-\cos(2x))\div2$$

anonymous · Answer

yeah

anonymous · Answer

ok thx