Recursive equation for an integral

[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

 

[StatusX]

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

 

[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)).

 

[VietDao29]

Uhmm, I just want to 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?

 

[MathematicalPhysicist]

it's n+(1/2)

 

[pizzasky]

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}.

 

[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.

 

[StatusX]

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)).

You should split the integral into the sum of two integrals, both of the form x^something/(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 I_m for a couple values of m.

 

[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})

 

[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))

 

[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

 

[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}

 

[StatusX]

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

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}}