# Integration

• Jul 11th 2010, 04:23 PM
mreazuddin
Integration
Hi All,
Can any one help me solve the following integration. I tried many times but I always get exactly the same thing I started with. It's like I = I. It would be very helpful if any one can shed some light on it.

cos(i*acos(x)) sin(j*acos(ax+b)) / sqrt(1-x^2)

i, j, a, b are constants. I tried integration by parts, with 'cos(i*acos(x)) / sqrt(1-x^2)' in one part and 'sin(j*acos(ax+b))' in another part. But I cannot solve it.

Thanks,
~Reaz
• Jul 12th 2010, 06:28 AM
Bwts
Make substitutions along the lines of:

$\displaystyle \sin (\arccos x) = \cos(\arcsin x) = \sqrt{1-x^2}$
• Jul 12th 2010, 07:32 AM
chiph588@
$\displaystyle \sin(j\arccos(ax+b))=j(ax+b)\sqrt{1-(ax+b)^2}$

And assuming $\displaystyle i\in\mathbb{N}$, $\displaystyle \cos(i\arccos(x))=T_i(x)$, where $\displaystyle T_i(x)$ is The Chebyshev polynomials of the first kind.
• Jul 12th 2010, 07:54 PM
mreazuddin
Hi,
Thanks for coming up.

I know $\displaystyle \sin(\arccos(x))=\sqrt(1-x^2)$. But I don't know why $\displaystyle \sin(j\arccos(ax+b))=j(ax+b)\sqrt(1+(ax+b)^2)$. And I actually replaced the Chebyshev's polynomial with it's equivalent so that I can integrate the product. Following is what I tried-

$\displaystyle \int \frac{\cos(i\arccos(x))\sin(j\arccos(ax+b))}{\sqrt {1-x^2}}$

$\displaystyle I = \sin(j\arccos(ax+b)) \int \frac{\cos(i\arccos(x))}{\sqrt{1-x^2}} - \int \frac{d}{dx}\sin(j\arccos(ax+b)) \int \frac{\cos(i\arccos(x))}{\sqrt{1-x^2}}$

$\displaystyle =-\frac{\sin(j\arccos(ax+b))\sin(i\arccos(x))}{i} - \frac{ja}{i} \int \frac{\cos(j\arccos(ax+b))\sin(i\arccos(x))}{\sqrt {1-(ax+b)^2}}$

$\displaystyle I = K - \frac{ja}{i} I2$

Now,
$\displaystyle I2 = \int \frac{\cos(j\arccos(ax+b))\sin(i\arccos(x))}{\sqrt {1-(ax+b)^2}}$

$\displaystyle =\sin(i\arccos(x)) \int \frac{\cos(j\arccos(ax+b))}{\sqrt{1-(ax+b)^2}} - \int \frac{d}{dx}\sin( i\arccos(x)) \int \frac{\cos(j\arccos(ax+b) )}{\sqrt{1-(ax+b)^2}}$

$\displaystyle = -\frac{\sin(j\arccos(ax+b))\sin(i\arccos(x))}{ja} - \frac{i}{ja} \int \frac{\cos(i\arccos(x))\sin(j\arccos(ax+b))}{\sqrt {1-x^2}}$

$\displaystyle I2 =\frac{i}{ja} K - \frac{i}{ja} I$

From the two equations of I and I2 we get I = I.

• Jul 12th 2010, 09:52 PM
chiph588@
I don't think this can be solved in general. See here.
• Jul 13th 2010, 12:00 PM
mreazuddin
Thanks. That also helps.
If I know it doesn't have a closed form then I can go for numerical methods.
• Jul 13th 2010, 12:50 PM
chiph588@
Quote:

Originally Posted by mreazuddin
Thanks. That also helps.
If I know it doesn't have a closed form then I can go for numerical methods.

What is this for anyway?
• Jul 25th 2010, 01:55 PM
mreazuddin
I was trying to compute double integral of product of Chebyshev's polynomial, with y dependent on x. And got this complex integral.
Thanks anyway.