Integral from Kyoto University Entrance Test

Wow,, this is one of a great integral for university entrance test…
I wanna post this because my calculus class just discussed the same kind of the problem

For 0<x<1, let f(x)=\int_{0}^{x} \frac{1}{\sqrt{1-t^2}}dt

  • Find \frac{d}{dx}f(\sqrt{1-x^2})
  • Find f(\frac{1}{\sqrt{2}})
  • Prove that f(x) + f(\sqrt{1-x^2})=\frac{\pi}{2}

I think the solution will be like this…
We have one foundation in calculus:: \frac{d}{dx} \int_c^x f(t)dt = f(x)
As f(x)=\int_{0}^{x} \frac{1}{\sqrt{1-t^2}}dt we know that

f(\sqrt{1-x^2})=\int_{0}^{\sqrt{1-x^2}} \frac{1}{\sqrt{1-t^2}}dt

Following the chain rule,, we have (I change \frac{d}{dx} into D_x)
D_x[f(\sqrt{1-x^2})]= D_{\sqrt{1-x^2}} [\int_{0}^{\sqrt{1-x^2}} \frac{1}{\sqrt{1-t^2}}dt] D_x(\sqrt{1-x^2})

D_x[f(\sqrt{1-x^2})]= \frac{1}{\sqrt{1-(1-x^2)}} \frac{-x}{\sqrt{1-x^2}}

D_x[f(\sqrt{1-x^2})]= \frac{-1}{\sqrt{1-x^2}}

Following the foundation of calculus,, we can take
f(\frac{1}{\sqrt{2}}) = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-t^2}}dt
We can substitute t=\sin{\theta}\rightarrow dt=\cos{\theta}d{\theta}
As \sin{\frac{\pi}{4}}=\frac{1}{\sqrt2} and \frac{1}{\sqrt{1-t^2}}=\frac{1}{\cos{\theta}}
So,, the integration will be like this::
f(\frac{1}{\sqrt{2}}) = \int_{0}^{\frac{\pi}{4}} \frac{\cos{\theta}}{\cos{\theta}}d{\theta}
f(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}

I actually don’t know how to generalize this,, but I found something unique…
As x=\frac{1}{\sqrt{2}} we have \sqrt{1-x^2}=\frac{1}{\sqrt{2}} too
So,, I think we can prove it by put x=\frac{1}{\sqrt{2}}
We have f(\frac{1}{\sqrt{2}}) + f(\frac{1}{\sqrt{2}}) = \frac{\pi}{2}
Sorry for this bad prove:mrgreen:

[As information,, Kyoto University is the 2nd best university in Japan (after Tokyo University). But I think Kyoto University has more peaceful place to study]

About mhar_teens

I'm only an ordinary boy... still confused about my future,, I only know that God has a great plan for me View all posts by mhar_teens

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