# Compute $\int \limits_{\cos x}^{\sin x} \sqrt{1 - t^2}\mathrm dt$

Problem: $\displaystyle \frac{d}{dx}\int \limits_{\cos x}^{\sin x} \sqrt{1 - t^2} \mathrm dt$ when $0<x<\pi$.

My attempt, see images.

Two questions:

1. Do I have to break it up into two cases?
2. I'm wrong. Correct answer is $1$. (Probably it is $\sin^2x+\cos^2x$). Why am I wrong?
– user61527
Nov 3, 2013 at 20:59
• @T.Bongers I thought 3 images was easier to read on dropbox than here. Nov 3, 2013 at 21:00

Using the Fundamental Theorem of Calculus, and the Chain Rule, differentiate the function with respect to $x$.

Before simplifying, we get $\cos x\sqrt{1-\sin^2 x}-(-\sin x)\sqrt{1-\cos^2 x}$.

Two cases are useful, since for $\pi/2\lt x\lt \pi$ we have $\sqrt{1-\sin^2 x}=|\cos x|$.

You don't need to compute the integral. Let's try a more general approach. Suppose $h$ is a function defined on some interval $[a,b]$ and that $f$ is another function taking its values in the interval $[a,b]$. Under these hypotheses, for any $c\in[a,b]$ and any $x$ in the domain of $f$, the integral

$$I(x)=\int_{c}^{f(x)} h(t)\,dt$$

is well defined. I'll assume also that $h$ is continuous and $f$ is differentiable; then, if $H$ is an antiderivative of $h$ (it exists because $h$ is continuous), we can write, by the fundamental theorem of calculus, $$I(x)=H(f(x))-H(c)$$ and so $$I'(x)=H'(f(x))f'(x)=h(f(x))f'(x)$$ again by the fundamental theorem of calculus ($H'=h$) and the chain rule. The choice of $c$ has effect on the value of $I(x)$, but not on the value of $I'(x)$.

Your integral can be written as $$F(x)=\int_{\cos x}^{\sin x}\sqrt{1-t^2}\,dt= \int_{0}^{\sin x}\sqrt{1-t^2}\,dt- \int_{0}^{\cos x}\sqrt{1-t^2}\,dt$$ and so $$F'(x)=\sqrt{1-\sin^2x}\sin' x-\sqrt{1-\cos^2x}\cos'x= |\cos x|\cos x+|\sin x|\sin x$$