# Problem involving $\int_{0}^{1}(\sin^{-1}x)^ndx$ and $\int_{0}^{1}(\cos^{-1}x)^ndx$

Let $A _{n}=\int_{0}^{1}(\sin^{-1}x)^ndx$ and $B_{n} = \int_{0}^{1}(\cos^{-1}x)^ndx$ for nonnegative integers n.

Prove that $A_{n} = \left ( \frac{\pi}{2} \right )^n - nB_{n-1}$ and $B_{n} = nA_{n-1}$

This is what i did for $A_{n}$, but it's hard for me to proceed further.

$A_{n} = \int_{0}^{1}(sin^{-1}x)^ndx \\ ~~~~\>=\left [ x(sin^{-1}x)^n) \right ]_{0}^{1} -n\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}(\sin^{-1}x)^{n-1}dx \\ ~~~~\>= \left ( \frac{\pi}{2} \right )^n -n\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}(\sin^{-1}x)^{n-1}dx$

Any ideas?

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Substitute $y=sin^{-1}x$ –  Theorem Nov 15 '12 at 10:33
Hey sorry, it's a typo. It's simply "=" . I edited it already. –  uohzxela Nov 15 '12 at 10:44

Note that $$A_n:=\int_{0}^{\frac{\pi}{2}}t^n\cos t dt\text{ and }B_n:=\int_{0}^{\frac{\pi}{2}}t^n\sin t dt.$$ Then, $$A_n+iB_n=\int_{0}^{\frac{\pi}{2}}t^n e^{it} dt,$$ and proceed by induction.

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Hey thanks for the answer but I don't quite get how you arrive that these steps. Can you explain to me how you get these? –  uohzxela Nov 15 '12 at 11:10

To evaluate $A_n$ use the substitution $x= \sin u$

To evaluate $B_n$ use the substitution $x= \cos u$

Now its easy to make reduction formulae for $u^n \cos u$ and $u^n \sin u$

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First, $x=\sin(u)$ yields $$A_n=\int_0^1\left(\sin^{-1}(x)\right)^n\mathrm{d}x=\int_0^{\pi/2}u^n\cos(u)\,\mathrm{d}u$$ and $x=\cos(u)$ gives $$B_n=\int_0^1\left(\cos^{-1}(x)\right)^n\mathrm{d}x=\int_0^{\pi/2}u^n\sin(u)\,\mathrm{d}u$$ Therefore, integration by parts yields \begin{align} A_n+iB_n &=\int_0^{\pi/2}u^ne^{iu}\,\mathrm{d}u\\ &=-iu^ne^{iu}{\Large]}_0^{\pi/2}+i\int_0^{\pi/2}nu^{n-1}e^{iu}\,\mathrm{d}u\\ &=\left(\frac\pi2\right)^n+in(A_{n-1}+iB_{n-1}) \end{align} Equate the real and imaginary parts.

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How did the imaginary part come about? I don't quite catch the intuition behind it. –  uohzxela Nov 15 '12 at 13:54
The imaginary part was introduced because $e^{iu}=\cos(u)+i\sin(u)$. –  robjohn Nov 15 '12 at 14:25