Analytical Solution of Double Integral in Symbolic Math

The following integral is known to have an analytical solution

int(sin(x)*(a^2*cos(x)^2 + sin(x)^2/a)^(1/2), x)

but Symbolic Math only returns the same command when I try to calculate it between the limits: 0 to pi.

Is there a way to calculate this integral analytically?

[David Goodmanson]

Hi Saeid, is this homework, and do you know if a>1 or a<1?

[Saeid]

Hi,

I'm actually a bit too old to try to figure out my homework solutions from a forum, but to answer your question: no this is not homework.

I am just wondering why Symbolic Math fails to yield an analytical solution that I assumed is relatively simple for it to produce.

"a" can have any value that is larger than 0.

For a>1 the result is a function containing arctan(sqrt(x^3-1))

For a<1 the result is a function containing arctanh(sqrt(1-x^3))

[David Goodmanson]

Hi Saeid,

I don't much like to say it, but Matlab symbolics is not real capable compared to some other stuff out there. Mathworks does not appear to be too concerned about it. I believe this is because they are after all a commercial entity, and the real money lies elsewhere. Not a great look, but it's hard to blame them.

For this problem you can substitute u = cos(x), du = -sin(x) dx, (1-u^2) = sin(x)^2. Since there are only even powers of sin and cos, integral{0,p1} dx = 2*integral{0,1} du. Saving the minus sign until some later time, you get

% int(sin(x)*(a^2*cos(x)^2 + sin(x)^2/a)^(1/2), x)
syms u a
I1 = int((a^2*u^2 + (1-u^2)/a)^(1/2), u)       % doesn't work

but with a little more help

I2 = int(((a^2-1/a)*u^2 + (1/a))^(1/2), u)

Assumptions help a bit.

assume(a<1)
I1 = int((a^2*u^2 + (1-u^2)/a)^(1/2), u)       % still doesn't work
I2 = int(((a^2-1/a)*u^2 + (1/a))^(1/2), u)

assume(a>1)
I1 = int((a^2*u^2 + (1-u^2)/a)^(1/2), u)         % both work and give the same result
I2 = int(((a^2-1/a)*u^2 + (1/a))^(1/2), u)

Integral I1 only works for a>1 in which case it gives the same result as I2. So consider just I2. For the definite integral

J2 = 2*int(((a^2-1/a)*u^2 + (1/a))^(1/2), u,0,1)

there are three cases above and of these only only the a>1 case works.

J2 = a   + log((a^3 - 1)^(1/2) + a^(3/2))/(a*(a^2 - 1/a)^(1/2))

That's it. This does evaluate to 2 as a-->1, as it should.

By all rights the a<1 case should have worked, but given the form it is in, it's a complex function that happens to come out real. So it's hard to see the inverse trig function in there.

In a fairly simple case like this, and given all the messing around one has to do getting syms to cooperate, for me personally it would be faster to rearrange it a bit and go to a table of integrals.

p.s. my fond school days are long gone.

[Saeid]

Thanks for the comprehensive response, David.

This definitely will help, and as you mentioned, I guess MATLAB has its focus on other areas where it does have a clear advantage, and almost no day goes by without me being surprised by a really useful feature I find in it!

Answer 1

[VBBV]

syms x 
a = 2; 
sol = int(sin(x).*(a.^2*cos(x).^2 + sin(x).^2/a).^(1/2),x,[0,pi])

sol = 

vpa(sol) % use vpa

ans = 

2.4543553915615044795171783624435

[VBBV]

When int is not able to calculate the definite integral for expression, you can use vpa to approximate the value. Otherwise try with integral for numerical integration

[VBBV]

As the given expression seems to be nonlinear, a closed form symbolic solution cannot exist for it between 0 and pi, You can instead use integrateByParts option to evaluate and simplify it.

syms x t
a = 2; 
sol = int(sin(x).*(a.^2*cos(x).^2 + sin(x).^2/a).^(1/2),x,[0 pi/2])

sol = 

F = integrateByParts(sol,sin(x))  % 

F = 

simplify(F)

ans =