Reduction of order on a 2nd order LI DE The equation and $y_1(x)$ are $$y_1(x)=x^{\frac{-1}{2}}\cos(x) $$
$$x^2y''+xy'+(x^2-\frac{1}{4})y=0$$
Using the formula that solves for $y_2(x)$
(the one with an integration factor over $y_1(x)$ squared, all inside an integral multiplied by $y_1(x)$) 
We get $$ x^{-1/2}\cos(x) \int \frac{\frac{1}{x}}{x^{1/4}\cos^2(x)}dx $$
Now from here what I thought to do  was to rationalize the integrand. There by making the equation easier for integration.
Just focusing on the integral I got 
$$\int \frac{1}{x^4\sqrt{x} *cos^2(x)} dx$$ 
Now it only seems to me that the integration has only gotten more complicated. Could this be done by parts or u-sub?
 A: We are given
$$x^2y''+xy'+\left(x^2-\dfrac{1}{4}\right)y=0 \tag 1$$
We are also given a solution as
$$y_1(x)=x^{-1/2}\cos(x)$$
Using Reduction of Order, a second solution is given by
$$y_2(x) = y_1(x) v(x) = x^{-1/2} \cos(x) v(x)$$
The first derivative is
$$y_2'(x) = \frac{\cos (x) v'(x)}{\sqrt{x}}-\frac{v(x) \cos (x)}{2 x^{3/2}}-\frac{v(x) \sin (x)}{\sqrt{x}}$$
The second derivative is
$$y_2''(x) = -\frac{\cos (x) v'(x)}{x^{3/2}}-\frac{2 \sin (x) v'(x)}{\sqrt{x}}+\frac{\cos (x) v''(x)}{\sqrt{x}}+\frac{v(x) \sin (x)}{x^{3/2}}+\frac{3 v(x) \cos (x)}{4 x^{5/2}}-\frac{v(x) \cos (x)}{\sqrt{x}}$$
Substituting this into the ODE
$x^2 y_2'' + x y_2'+\left(x^2-\dfrac{1}{4}\right)y_2$
$= x^2\left(-\dfrac{\cos (x) v'(x)}{x^{3/2}}-\dfrac{2 \sin (x) v'(x)}{\sqrt{x}}+\dfrac{\cos (x) v''(x)}{\sqrt{x}}+\dfrac{v(x) \sin (x)}{x^{3/2}}+\dfrac{3 v(x) \cos (x)}{4 x^{5/2}}-\dfrac{v(x) \cos (x)}{\sqrt{x}}\right) + x \left(\dfrac{\cos (x) v'(x)}{\sqrt{x}}-\dfrac{v(x) \cos (x)}{2 x^{3/2}}-\dfrac{v(x) \sin (x)}{\sqrt{x}}\right) + \left(x^2-\dfrac{1}{4}\right)x^{-1/2} \cos(x) v(x) = 0$
Simplifying, we get
$$x^{3/2} \left(\cos (x) v''(x)-2 \sin (x) v'(x)\right) = 0$$
Now solve for $v(x)$ and then find $y_2(x)$.
Can you proceed?
The final result should be 
$$y_2(x) = \dfrac{\cos (x) (c_1 \tan (x)+c_2)}{\sqrt{x}}$$
You can substitute this into $(1)$ and verify it satisfies the ODE.
