Evaluate the following integral.

Question

I tried solving the question using substitution, but I always seem to end up with a recursive formula. Any hints about how to do the question?

$$\int_{0}^{1} \frac{x^{2n+1}}{\sqrt{1-x^2}} dx.$$

On this post, there is a clean answer :) – Tak Feb 28 '16 at 17:34 — Tak, Feb 28 '16 at 17:34

score 2 · Accepted Answer · answered Feb 22 '16 at 07:50

2

HINT:

Set $x=\sin y$ to get

$$\int_0^{\pi/2}\sin^{2n+1}y\ dy=I_{2n+1}$$

Form this, $$I_m=\dfrac{m-1}mI_{m-2}$$

answered Feb 22 '16 at 07:50

lab bhattacharjee

score 0 · Answer 2 · answered Feb 22 '16 at 08:13

0

Hint:

Set $x=\sin (u)$ and $dx=\cos (u)du$ then $\sqrt{1-x^2}=\sqrt{1-\sin^2 (u)}=\cos(u)$ and $u=\arcsin(x)$

So

$$\int\sin^{2n+1}(u)du$$

answered Feb 22 '16 at 08:13

3SAT

2 Answers2