Math Help - how to integrate

1. how to integrate

Hi,

How to integrate x^2/(xsinx + cosx)^2.

I am clueless. It does not fit into any formula straight away. If I want to try integration by parts, it appears to be lengthy and i do not know whether it will lead to the answer. How to go about this? do we have any particular method for solving such type of sums.

whether integration within integration is permissible. If so, when it is applicable / possible.

Any guidance welcome.

with warm regards,

ARANGA

2. Re: how to integrate

Integration by parts is most likely to give you the answer. I don't understand what you mean by integration within integration. During integration by parts, I suppose you can consider that a sort of "integration within integration", but I am not sure where you are going with that.

3. Re: how to integrate

I can't seem to find the right u and v to integrate by parts. There are four factors: two x's and two $\frac{1}{x\sin{x}+\cos{x}}$'s. I don't see any way of integrating something with $x\sin{x}+\cos{x}$ or $(x\sin{x}+\cos{x})^2$ in the denominator, so I'm left with $dv=dx$, $dv=x\,dx$, or $dv=x^2\,dx$. But they all give me $(x\sin{x}+\cos{x})^3$ in the denominator, so it seems like I'm going the wrong way.

- Hollywood

4. Re: how to integrate

One way of doing it, which is a bit on the cheesy side, is to try and get the integrand into a quotient derivative form.

$h(x)=\dfrac{f(x)}{g(x)}$

$\dfrac{dh}{dx}=\dfrac{f'g - fg'}{g^2}$

$g(x)=x \sin(x)+\cos(x)$

$g'(x)= \sin(x) +x \cos(x) - \sin(x) = x \cos(x)$

$x^2 = f'(x)\left(x \sin(x) + \cos(x)\right) - f(x)\left(x \cos(x)\right)$

you can see you want $f(x)$ to have a $-x \cos(x)$ term in it, and with some toying around you can find that

$f(x) = \sin(x) - x\cos(x)$

and thus

$h(x) = \dfrac{f(x)}{g(x)} = \dfrac{\sin(x)-x\cos(x)}{\cos(x)+x\sin(s)}+C$

But this only works for this very contrived problem. Solving the general differential equation above is by no means guaranteed to be possible.

5. Re: how to integrate

Here is how I would start:

$u = x\sec x$, $dv = \dfrac{x\cos x}{(x\sin x + \cos x)^2}dx$

Then $du = \sec x(x\tan x + 1)dx$

$\displaystyle \int \dfrac{x^2}{(x\sin x + \cos x)^2}dx = -\dfrac{x\sec x}{x\sin x + \cos x} + \int \dfrac{\sec x(x\tan x+1)}{x\sin x + \cos x}dx$

This reduces the exponent of the denominator. I don't know if it solves the problem, but it might.