Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

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 $u^n \sin u$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.