# Thread: calculate the derivative of an integral

1. ## calculate the derivative of an integral

d/dx $\displaystyle \int_x^0 \! sin^2(t) \, \mathrm{d}x.$

so first I flipped it & changed the sign to get the x on top:
d/dx - $\displaystyle \int_0^x \! sin^2(t) \, \mathrm{d}x.$

Now my question is, what is the integral of sin^2(t)?

Like I know for just integral sinxdx=-cosx...

If I set cos2x=1-sin^2(x)..will this work?

2. First, recognize that a function of t is independent of x and is therefore a constant with respect to x, this turns the integral into:

$\displaystyle sin^2(t)\frac{d}{dx} \int_x^0 ~dx$

Now, solve the integral, which is just x evaluated from 0 to x:

$\displaystyle sin^2(t)\frac{d}{dx} x|^0_x$

$\displaystyle sin^2(t)\frac{d}{dx} -x$

Now we take the derivative, which is just -1:

$\displaystyle \frac{d}{dx}\int_x^0 sin^2(t) ~dx = -sin^2(t)$

3. But if you meant $\displaystyle \frac{d}{dx}\int_x^0 sin^2(t)dt= -\frac{d}{dx}\int_0^x sin^2(t)dt$ ("dt", not "dx")
then you could integrate by using the trig identity "$\displaystyle sin^2(t)= \frac{1}{2}(1- cos(2t))$".

However integrating at all misses the point of this problem- it is an exercise in the "Fundamental Theorem of Calculus":
$\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$.

Do it both ways as a check!

4. ok, thank you so much! that made so much more sense than my book!