# Recursive equation for an integral

1. May 29, 2006

### MathematicalPhysicist

i have:
$$I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx$$

and i need to find a recursive equation for the above integral, and this is what i got so far:
$$I_n=\left[x^{n+1/2}*arcsin(x)\right]_{0}^{1}+(n+1/2)\int_{0}^{1}x^{n-1/2}arcsin(x)dx$$

Last edited: May 29, 2006
2. May 29, 2006

### StatusX

Try multiplying the top and bottom of the original integrand by 1-x2.

3. May 29, 2006

### MathematicalPhysicist

it still doesn't work, because i get an integral with x^(n+3/2)/(sqrt(1-x^2))^3, where i should get an integral of x^(n+3/2)/(sqrt(1-x^2)).

4. May 29, 2006

### VietDao29

Uhmm, I just wanna ask, is it (n + 1) / 2 or n + (1 / 2)?
Is it:
$$I_n=\int_{0}^{1} \frac{x^{n+ \frac{1}{2}}}{\sqrt{1-x^2}}dx$$
or
$$I_n=\int_{0}^{1} \frac{x^{\frac{n + 1}{2}}}{\sqrt{1-x^2}}dx$$?

5. May 29, 2006

### MathematicalPhysicist

it's n+(1/2)

6. May 29, 2006

HINT: Consider $$\frac{d}{dx}[x^{n+\frac{1}{2}}\sqrt{1-x^2}]$$.

Once you have obtained the recursive equation, you will just need to modify it slightly to obtain $$I_{n}$$ in terms of $$I_{n-2}$$.

Last edited: May 29, 2006
7. May 29, 2006

### VietDao29

You can use StatusX's hint to solve your problem. And also you should choose u, and v wisely.
$$I_n = \int_0 ^ 1 \frac{x ^ {n + \frac{1}{2}}}{\sqrt{1 - x ^ 2}} dx = \int_0 ^ 1 \frac{x ^ {n + \frac{1}{2}} (1 + x ^ 2)}{\sqrt{(1 - x ^ 2) ^ 3}} dx = - \frac{1}{2} \int_0 ^ 1 \frac{x ^ {n - \frac{1}{2}} (1 + x ^ 2)}{\sqrt{(1 - x ^ 2) ^ 3}} d(1 - x ^ 2)$$
$$= \int_0 ^ 1 x ^ {n - \frac{1}{2}} (1 + x ^ 2) d \left( \frac{1}{\sqrt{1 - x ^ 2}} \right)$$
Now let $$u = x ^ {n - \frac{1}{2}} (1 + x ^ 2)$$, and $$dv = d \left( \frac{1}{\sqrt{1 - x ^ 2}} \right)$$.
Can you go from here? :)
Recheck my maths if possible. I'm having a slight headache.

8. May 30, 2006

### StatusX

You should split the integral into the sum of two integrals, both of the form xsomething/(1-x2)3/2. These can both be integrated by parts to get two boundary terms (which must be combined before being evaluated), and multiples of Im for a couple values of m.

Last edited: May 30, 2006
9. May 30, 2006

### benorin

Use x=sin(u)

Use the substitution $$x=\sin u$$ so that $$dx=\cos u du$$ to get

$$I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u du$$​

The recurrence you are looking for is $$I_{n+2}=f(n)I_{n}$$
where f(n) is a rational function of n.

P.S. (the one you are not looking for is $$I_{n}I_{n+1}=\frac{\pi}{2n+3}$$)

10. May 30, 2006

### MathematicalPhysicist

benorin, i tried your approach but still didn't get a recursive eq, here what i got:
$$I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} (t) dt=\left[tsin^{n+1/2}(t)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(n+1/2)sin^{n-1/2}(t)d(sin(t))$$

Last edited: May 30, 2006
11. May 30, 2006

### MathematicalPhysicist

and statusx, here what i got from your appraoch:
$$I_n=\int_{0}^{1} \frac{x^{n+1/2}}{\sqrt{1-x^2}}dx=\left[\frac{x^{n+3/2}}{(n+3/2)(\sqrt{1-x^2})^{3}}\right]_{0}^{1}-\int_{0}^{1}\frac{3x^{n+5/2}}{(n+3/2)(\sqrt{1-x^2})^{5}}dx-\left[\frac{x^{2n+2}}{(2n+2)(\sqrt{1-x^2})^{3}}\right]_{0}^{1}+\int_{0}^{1}3/(2n+2)\frac{x^{2n+3}}{(\sqrt{1-x^2})^{5}}dx$$

Last edited: May 30, 2006
12. May 30, 2006

### benorin

Use a trick

$$I_{n+2}=\int_{0}^{\frac{\pi}{2}} \sin ^{n+2+1/2} u du$$
$$=\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u\sin ^{2} u du$$
$$= \int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u(1-\cos ^{2} u) du$$
$$= \int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u du -\int_{0}^{\frac{\pi}{2}} \sin ^{n+1/2} u\cos ^{2} u du$$
$$= I_{n}-\int_{0}^{\frac{\pi}{2}} \left( \sin ^{n+1/2} u\cos u\right) \cos u du$$
integration by parts (in the last integral): put

$$U=\cos u, dU = -\sin u, dV=\sin ^{n+1/2} u\cos u du, V=\frac{1}{n+1+1/2}\sin ^{n+1+1/2} u$$

to get

$$I_{n+2}= I_{n}-\int_{0}^{\frac{\pi}{2}} \left( \sin ^{n+1/2} u\cos u\right) \cos u du$$
$$= I_{n}-\left[ \frac{1}{n+2+1/2}\cos u\sin ^{n+2+1/2} u \right] _{u=0}^{\frac{\pi}{2}} - \frac{1}{n+1+1/2}\int_{0}^{\frac{\pi}{2}} \sin ^{n+2+1/2} u du$$
$$=I_{n}- \frac{1}{n+1+1/2}I_{n+2}$$

rewritten this becomes $$I_{n+2}=I_{n}- \frac{2}{2n+3}I_{n+2}$$

which gives what your after, namely $$I_{n+2}= \frac{2n+3}{2n+5}I_{n}$$

Last edited: May 30, 2006
13. May 30, 2006

### StatusX

I think you're taking the wrong u and dv. Remember:

$$\int \frac{x}{(1-x^2)^{3/2}}dx = \frac{1}{\sqrt{1-x^2}}$$