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.
Best Answer
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}