Fundamental theorem of calculus question with trig limits Can anyone help me answer this question using the fundamental theorem of calculus. 
Take the integral of the following integral. 
$$\int_{\cos(x)}^{\sin(x)}\sqrt{1-t^2}\,\mathrm dt$$
For when $0 < x < \pi/2$
My thought to solve it is to split up the integral into two integrals so we have two integrals with a function as the upper limit. Can someone show me the steps of how it could be done? Thank you. 
Please ask if something was unclear, I have asked a similar question earlier. But now I wanted to specify the usage of "the fundamental theorem of calculus" in the way of solving it. 
 A: hint
$t\mapsto \sqrt {1-t^2} $ is continuous at $[0,1]  \implies $
by FTC,
$F:x\mapsto \int_{\cos (x)}^{\sin (x)} \sqrt{1-t^2}dt $ is differentiable at $(0,\frac {\pi}{2}) $ and
$$F'(x)=\sqrt {1-\sin^2 (x)}\cos (x)+\sqrt {1-\cos^2 (x)}\sin (x)=1$$
but
$$F (\frac {\pi}{4})=0$$
thus
$$\int_{\cos(x)}^{\sin(x)}\sqrt {1-t^2}dt=x -\frac {\pi}{4}.$$
A: The first step is to find the integral
$$
\int \sqrt{1-t^2}dt
$$
this is classical indefinite integral that can be evaluated using the trigonometric substitution 
$t=\sin u\quad$ that implies $ \qquad dt=\cos u\; du \quad$  and$ \qquad \sqrt{1-t^2}=\cos u$
If you solve this you find the primitive:
$$
y=\frac{1}{2}\left( \arcsin t +t\sqrt{1-t^2} \right)
$$
Now you can evaluate the definite integral substituting the limits:
$$
\left[\frac{1}{2}\left( \arcsin (\sin x) +\sin x\sqrt{1-\sin^2 x} \right)\right]-\left[\frac{1}{2}\left( \arcsin (\cos x) +\cos x\sqrt{1-\cos^2 x} \right)\right]=x-\frac{\pi}{4}
$$
and you can conclude evaluating
$$
\int_0^{\frac{\pi}{2}}(x-\frac{\pi}{4})dx=0
$$
A: Hint 
By fundamental theorem of calculus if $$G(x)= \int_{g(x)}^{f(x)} h(t) dt$$ then $$G'(x)=f'(x)h(f(x))-g'(x)h(g(x))$$
Using that $$F'(x)=1$$ which is constant hence $F(x)$ is linear.  And also notice that $F(\frac {\pi}{4})=0$
