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need your help on this:

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?

  • 0
    Substitute $y=sin^{-1}x$2012-11-15
  • 0
    Hey sorry, it's a typo. It's simply "=" . I edited it already.2012-11-15

3 Answers 3