Changing the bounds of integration I have a question that asks me to find the derivative of this integral, with out evaluation the intergral.
$$\int_{\sin x}^{\cos x}\frac {1}{1-t^2}dt$$
I think I need to use U-substitution and the chain rule, but I cant figure out how to apply it in this case.
Any hints would be appreciated.
 A: As carmichael561 commented, the  fundamental theorem of calculus tells that $$\frac d {dx}\int_{a(x)}^{b(x)} f(t) \, dt=f(b(x)) b'(x)-f(a(x)) a'(x)$$ For your case $$a(x)=\sin(x)\implies a'(x)=\cos(x)$$ $$b(x)=\cos(x)\implies b'(x)=-\sin(x)$$ $$f(b(x))=\frac 1{1-\cos^2(x)}=\frac{1} {\sin^2(x)}$$ $$f(a(x))=\frac 1{1-\sin^2(x)}=\frac{1} {\cos^2(x)}$$
Just finish.
A: Only elementary calculus is required. Note that I prefer using differentials(which is basically an implicit u substitution).
Recall the fundamental theorem of calculus, for a constant $a$
\begin{equation}
\frac{d}{dx} \int_a^x f(t) dt = f(x)
\end{equation}
so
\begin{align}
\frac{d}{dx}\int_{\sin x}^{\cos x} \frac{1}{1 - t^2} dt &= \frac{d}{dx}\int_0^{\cos x} \frac{dt}{1 - t^2} - \frac{d}{dx} \int_0^{\sin x} \frac{dt}{1 - t^2}\\
&= \frac{d(\cos x)}{dx} \frac{d}{d(\cos x)}\int_0^{\cos x} \frac{dt}{1 - t^2} - \frac{d(\sin x)}{dx} \frac{d}{d(\sin x)} \int_0^{\sin x} \frac{dt}{1 - t^2}\\
&= \Big(-\sin x\Big)\Big(\frac{1}{1 - \cos^2 x}\Big)  - (\cos x)\Big( \frac{1}{1 - \sin^2 x}\Big)\\
&= \frac{-\sin x}{\sin^2 x} - \frac{\cos x}{\cos^2 x}\\
&= -\Big(\frac{1}{\sin x} + \frac{1}{\cos x}\Big)\\
&= - \csc x - \sec x
\end{align}
