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Consider

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)^n\mathrm dx=f(n)\tag1$$ $n\ge0$

Using wolfram integrator, it gives us a closed form for $f(n)$, we managed to identify a pattern as shown below.

setting $n=1,2,3,4$ and $5$

We notices the closed from has a pattern that involved binomial coefficents.

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)\mathrm dx=2!-1!\cdot{\pi\over 2}$$

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)^2\mathrm dx=4!-3!\cdot{\pi\over 2}-1!\cdot{3\pi^2\over 4}$$

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)^3\mathrm dx=6!-5!\cdot{3\pi\over 2}-4!\cdot{3\pi^2\over 4}+3!\cdot{\pi^3\over 8}$$

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)^4\mathrm dx=8!-7!\cdot{4\pi\over 2}-6!\cdot{6\pi^2\over 4}+5!\cdot{4\pi^3\over 8}+4!\cdot{\pi^4\over 16}$$

$$\int_{0}^{1}(\sin^{-1}x \cos^{-1}x)^5\mathrm dx=10!-9!\cdot{5\pi\over 2}-8!\cdot{10\pi^2\over 4}+7!\cdot{10\pi^3\over 8}+6!\cdot{5\pi^4\over 16}- 5!\cdot{\pi^5\over 32}$$

$$f(6)=12!-11!\cdot{6\pi\over 2}-10!\cdot{15\pi^2\over 4}+9!\cdot{20\pi^3\over 8}+8!\cdot{15\pi^4\over 16}- 7!\cdot{\pi^5\over 32}-6!\cdot{\pi^6\over 64}$$

We generalised

$$f(n)=\sum_{k=0}^{n}(-1)^{\lceil {k/2}\rceil}(2n-k)!{n\choose k}\cdot{\pi^k\over2^k}\tag2$$

How does one prove (2)?

An attempt:

$I$ becomes

$$\int_{0}^{1}\left[{\pi\over 2}\cos^{-1}x-(\cos^{-1}x)^2\right]^n\mathrm dx\tag3$$

probably using the binomial theorem to expand and integrate term by term, I guess

I am stuck at this point, no sure what to do

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  • $\begingroup$ Did my hint help you ? $\endgroup$ – Zaid Alyafeai Feb 11 '17 at 22:07
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Hint

First note that in (3)

$$\int_{0}^{1}\left[{\pi\over 2}\arccos x-(\arccos x)^2\right]^n\mathrm dx = \sum_{k=0}^n {n \choose k }\left(\frac{\pi}{2} \right)^{k} (-1)^{n-k} \int^1_0 (\arccos x)^{2n-k}\,dx $$ Solve the general formula

$$I = \int_{0}^{1}( \arccos x)^{n}dx$$

Let $x = \cos(t)$ then $dx =-\sin(t) \,dt$

$$I = \int_{0}^{\frac{\pi}{2}}t^n \sin(t) \,dt $$

Then you can easily find the integral using ingration by parts $n$ times.

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