Calculus – How to Find the Derivative of a Function

Question

$$f(x) = \int_0^{\cos x}t^2 \, dt$$

I computed and got $f'(x) = – \cos^2(x) \sin(x)$. I checked the answer in the book and I got this wrong. But I think I did this correctly. I used part 1 of FTC to do this. Any ideas?

Answer 1

Note that: $$\begin{align} f(x) &= \int_{a(x)}^{b(x)} g(t)\,dt = G\bigl(b(x)\bigr)-G\bigl(a(x)\bigr) \\ f'(x) &= g\bigl(b(x)\bigr)\cdot b'(x)-g\bigl(a(x)\bigr)\cdot a'(x) \end{align}$$ Hence: $$\begin{align} f(x) &= \int_0^{\cos x}t^2 \, dt=G(\cos x)-G(0) \\ f'(x)&=(\cos x)^2\cdot (\cos x)'-(0)^2\cdot 0' \\ &=\cos^2x \cdot (-\sin x) \\ &=-\cos^2x\cdot \sin x \end{align}$$