Fundamental theorem of calculus question with trig limits

Question

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.

Answer 1

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

$$\int_{\cos(x)}^{\sin(x)}\sqrt {1-t^2}dt=x -\frac {\pi}{4}.$$

Answer 2

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 $$ \int_0^{\frac{\pi}{2}}(x-\frac{\pi}{4})dx=0 $$

Answer 3

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 $F(\frac {\pi}{4})=0$