# $\int_3^5{\frac{x^2}{1+x^2}dx}$ by differentiation under the integral

I'm trying an easy problem to get my bearings using the method here.

The integral is $$\int_3^5{\frac{x^2}{1+x^2}dx}$$.

I would like to proceed, if possible to solve by defining:

$$F(y) = \int_3^5{\frac{\sin{(y\cdot x})}{1+x^2}dx}$$

Then obtaining $$-F''(y) = \int_3^5{\frac{x^2\sin{(y\cdot x})}{1+x^2}dx}$$

Adding gives $$F(y) - F''(y)= \int_3^5{\frac{(1+x^2)\sin{(y\cdot x})}{1+x^2}dx}$$

or $$F(y) - F''(y) - \frac{\cos{3y}-\cos{5y}}{y} = 0.$$

This is where I'm stuck. I don't know how to solve the differential equation. Any help would be greatly appreciated. I'm assuming that this can be done.

-
If you use cos instead of sin, then $-F''(y)$ will be just what you're looking for when $y=0$. Then, for $(\sin 3y - \sin 5y)/y$, you'd find the limit as $y\to0$. – Michael Hardy Sep 6 '11 at 0:15
@Michael Hardy: F(y) would still be part of the result, though, wouldn't it? I'm trying not to integrate the denominator at all if possible. My goal is to avoid integrating the fractions. – Matt Groff Sep 6 '11 at 0:36
@Michael Hardy: $F(0)$ must be involved as well. – anon Sep 6 '11 at 0:37
This is really roundabout, but doable. Using the original version with $\sin$, the result will have the form of $F$ in terms of an integral, but you need $F'(0)$ for your final answer so it disappears. One way is to solve $u-y''=\cos(kt)/t$ by introducing $v=u'$ and then writing the second-order DE as a first-order system ($\vec{q}=(u,v)^T$) $$q'=\begin{pmatrix}0&1\\1&0\end{pmatrix}q-{0\choose\cos(kt)/t}$$ which can be solved using an integrating factor with the matrix exponential. – anon Sep 6 '11 at 0:43
As anon mentions, I should say that calculating limits should be easy for what I anticipate - hence I can get F(0), F''(0), and so on... Thanks to anon, by the way. – Matt Groff Sep 6 '11 at 0:45

$y''-y+ \frac{\cos3x-\cos5x}{x}=0$ ,
This is nonhomogeneous second-order ordinary differential equation of the form: $y''+p(x)y'+q(x)y-g(x)=0$ ,which can be solved if the general solution to the homogenous version is known, in which case variation of parameters can be used to find the particular solution:
$y_p=-y_1(x)\int\frac{y_2(x)g(x)}{W(x)}\, dx + y_2(x)\int\frac{y_1(x)g(x)}{W(x)}\, dx$ , where $y_p$ is particular solution, $y_1(x)$ and $y_2(x)$ are the homogeneous solutions of the equation $y''+p(x)y'+q(x)y=0$ and W(x) is the Wronskian of these two functions.