d/dx
so first I flipped it & changed the sign to get the x on top:
d/dx -
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?
d/dx
so first I flipped it & changed the sign to get the x on top:
d/dx -
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?
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:
Now, solve the integral, which is just x evaluated from 0 to x:
Now we take the derivative, which is just -1:
But if you meant ("dt", not "dx")
then you could integrate by using the trig identity " ".
However integrating at all misses the point of this problem- it is an exercise in the "Fundamental Theorem of Calculus":
.
Do it both ways as a check!