Find the derivative of I(x) = $\int _{\sin\left(x\right)}^{\cos\left(x\right)}\arctan\left(t^2\right)\,dt$ $$I'(x)= \frac{d}{dx}\left(\int_{\sin\left(x\right)}^{\cos\left(x\right)}\arctan\left(t^2\right)\,dt\right)$$
I'm not sure how to approach this problem, initially I thought to use the Fundamental theorem with the substitution $u = \cos(x)$ but then I still can't use the Fundamental theorem because of the $\sin(x)$. A point in the right direction would be appreciated.
 A: Simply apply Newton-Leibnitz theorem, according to whichLet f(x, t) be a function such that the partial derivative of f with respect to t exists, and is continuous. Then, (click this link for formula). Hope this helps.
A: FTC $1$ States that if $f:[0,b]\to\mathbb{R}$ is integrable and $F(x)=\int_0^xf(t)dt$, then $F$ is diferentiable and $$F'(x)=f(x)$$
Now, we have $$I(x)=\int_0^{\cos x}\arctan(t^2)dt-\int_0^{\sin x}\arctan(t^2)dt$$
Using the Chain Rule and TFC $1$ we get:
\begin{align*}
I'(x)&=\arctan(\cos^2 x)\frac{d}{dx}(\cos x)-\arctan(\sin^2 x)\frac{d}{dx}(\sin x)\\
&=\arctan(\cos^2 x)\cdot(-\sin x)-\arctan(\sin^2 x)\cdot(\cos x)\\
&=-(\sin x)\arctan(\cos^2 x)-(\cos x)\arctan(\sin^2 x)
\end{align*}
A: Use Leibniz's rule:
$$b'(x) = (\cos x)' = -\sin x, a'(x) = (\sin x)' = \cos x$$
which gives you 
$$ I'(x) = - \sin x \arctan(\cos^2 x) - \cos x \arctan(\sin^2 x)$$
EDIT: I got the signs mixed up
A: just use it with chain rule
$$\begin{align}
\int_{b}^{a}f(x)dx&=F(b)-F(a)\\
I(x)&=\int_{\sin x}^{\cos x}\overbrace{\arctan(t^2)}^{f(t)}dt\\
I'(x)&=\frac{d}{dx}\int_{\sin x}^{\cos x}f(t)dt=\frac{d}{dx}[F(\cos x)-F(\sin x)]\\
&=(\cos x)'f(\cos x)-(\sin x)'f(\sin x)\\
&=-\sin x\arctan\cos^2x-\cos x\arctan\sin^2x
\end{align}$$
