Dependency on x? Definite integrals

Question

Prove that the values of the $$\int_{-cos(x)}^{sin(x)} \frac{1}{\sqrt{1-t^2}}dt, x \in (0, \frac{\pi}{2})$$

Do not depend on x.

I don't know what this means. I just found the derivative.

$$\leftrightarrow -\int_{0}^{-cos(x)} \frac{1}{\sqrt{1-t^2}}dt + \int_{0}^{sin(x)} \frac{1}{\sqrt{1-t^2}}dt$$

$$= - (sin(x))\frac{1}{\sqrt{1-cos^2 (x)}} + cos(x)\frac{1}{\sqrt{1-sin^2(x)}}$$

What now?

Hint: take the derivative with respect to $x$. Now think on what should happen to a function that does not depend on $x$ if you derive it with respect to $x$ — Alonso Delfín, Jun 09 '17 at 09:07

score 0 · Accepted Answer · answered Jun 09 '17 at 09:21

$$fx)=\int_{-cos(x)}^{sin(x)} \frac{1}{\sqrt{1-t^2}}dt, x \in (0, \frac{\pi}{2}) \\\to \\ f'= - (sin(x))\frac{1}{\sqrt{1-cos^2 (x)}} + cos(x)\frac{1}{\sqrt{1-sin^2(x)}}=\\ - (sin(x))\frac{1}{\sqrt{\sin ^2 (x)}} + cos(x)\frac{1}{\sqrt{\cos^2(x)}}=\\\dfrac{-\sin x}{|\sin x|}+\dfrac{\cos x}{|\cos x|}$$ when $x \in (0, \frac{\pi}{2}) \sin x>0 , \cos x >0$ so $$f'=\dfrac{-\sin x}{|\sin x|}+\dfrac{\cos x}{|\cos x|}=\\ \dfrac{-\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=-1+1=0 $$ $f'=0 \to f(x)=const$ so f(x) does not depend on $x$

score 0 · Answer 2 · answered Jun 09 '17 at 09:22

$\displaystyle \int_{-\cos(x)}^{\sin(x)} \frac{1}{\sqrt{1-t^2}}\,dt$

$=\sin^{-1}(\sin(x))-\sin^{-1}(-\cos(x))=$

$=\sin^{-1}(\sin(x))-\sin^{-1}(\sin\left(x-\frac{π}{2}\right)$

On this interval, this is equal to

$\displaystyle x-\left(x-\frac{π}{2}\right)=\boxed{\frac{π}{2}}$.

Dependency on x? Definite integrals

2 Answers2