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for the following Question, i had to prove this :

that for every $$-1\le y \le 1 \\ \arcsin(y) + \arccos(y) = \frac{\pi}{2}$$ NOTE: this I've shown this using basic trigonometric id's

and (probably) somehow use this to prove the following :

$$\int_0^{\sin^2(x)}\arcsin(\sqrt {t})dt + \int_0^{\cos^2(x)}\arccos(\sqrt {t})dt = \frac{\pi}{4} $$

I've been working quite some time on this one, and will appreciate your help on proving this, thank you.

doraemonpaul
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Simba
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3 Answers3

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Let $$f(x)=\int_0^{\sin^2(x)}\arcsin(\sqrt {t})dt + \int_0^{\cos^2(x)}\arccos(\sqrt {t})dt $$ then we can see that $$f'(x)=2\sin(x)\cos(x)\arcsin(\sin x)-2\sin(x)\cos(x)\arccos(\cos x)=0$$ so $$f(x)=f\left(\frac{\pi}{4}\right)=\int_0^{1/2}\arccos(\sqrt{t})dt+\int_0^{1/2}\arcsin(\sqrt{t})dt=\frac{1}{2}\times\frac{\pi}{2}=\frac{\pi}{4}$$

Pedro
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  • i am confused, how did you get the answer for the integrals?

    and am i allowed to assume that the derivitive is what you have wroten? why?

     – Simba Apr 27 '13 at 13:41
  • @eyal the derivative of this function defined by an integral: $\int_0^{g(x)}f(t)dt$ is $f(g(x))g'(x)$ and since we find the derivative zero so the function $f$ is constant and its value is any value taken by $f$ on some point which we choose carefuly to make calculus easy. –  Apr 27 '13 at 14:11
  • thats a great trick! am i allowed to assume that the dervitive of an integral, ia allways like this when the function is continous? what are the conditions? – Simba Apr 27 '13 at 15:42
  • this is an amaizing answer. thank you very much – Simba Apr 27 '13 at 15:50
  • @eyal you're welcome. –  Apr 27 '13 at 16:29
  • Deserves an UV too! – amWhy Jul 23 '14 at 12:08
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Let $u=\arcsin t, t=\sin u, dt =\cos u du\implies \dfrac{du}{dt}=\dfrac1{\cos u}=\dfrac1{\sqrt{1-t^2}}$

Similarly, $\dfrac{d\arccos t}{dt}=-\dfrac1{\sqrt{1-t^2}}$

So, $$\dfrac{d(\arccos t+\arcsin t)}{dt}=0$$

$$\implies \arccos t+\arcsin t=C$$ where $C$ is an arbitrary constant of indefinite integral.

Now, $C=\arccos 0+\arcsin 0=\dfrac\pi2+0$

Inceptio
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lab bhattacharjee
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Here is a proof

$$ \arcsin(y) + \arccos(y) = \frac{\pi}{2} \implies \sin( \arcsin(y) + \arccos(y) ) = 1 $$

$$ \implies y.y+\sqrt{1-y^2}. \sqrt{1-y^2} = 1 $$

$$ \implies 1=1. $$

Note: We used the identities

1)

$$ \sin(a+b)= \sin(a)\cos(b)+\sin(b)\cos(a), $$

$$ \sin(\arcsin(x))=x,\quad \sin(\arccos(x))=\sqrt{1-x^2}= \cos(\arcsin(x)).$$

Mhenni Benghorbal
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