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#### Ackbach

##### Indicium Physicus
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Originally posted by Srengam on 9/5/2011: integrate
$$\int[(1-2\cos(x))(3+2\cos(x))e^{\frac{1}{2}\,x+\cos(x)}]\,dx.$$
My solution:

One way to integrate involves solving a first-order linear ordinary differential equation. First, note that
$$\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^{2}}.$$
That, of course, is just the quotient rule for derivatives. You can integrate it once to obtain
$$\frac{u}{v}+C=\int\frac{vu'-uv'}{v^{2}}\,dx.$$
Now, if you could get the integrand to look like the integrand I just mentioned, you'd be done. Let's say you write
$$\int\frac{e^{-x/2-\cos(x)}(1-2\cos(x))(3+2\cos(x))}{e^{-x-2\cos(x)}}\,dx.\quad (1)$$
All I've done is write the exponential in the denominator, and then multiplied top and bottom by the new denominator, because I want to get a $v^{2}$ in the denominator. So now I want $v=e^{-x/2-\cos(x)}.$ This forces my quotient rule to look like this:
$$\frac{vu'-uv'}{v^{2}}=\frac{e^{-x/2-\cos(x)}u'-e^{-x/2-\cos(x)}(-1/2+\sin(x))u}{e^{-x-2\cos(x)}}.$$
Equating the numerator of this RHS with the previous numerator of (1) yields the first-order linear ordinary differential equation
$$u'-(-1/2+\sin(x))u=(1-2\cos(x))(3+2\cos(x)).$$
The solution to this DE is
$$u=e^{-x/2-\cos(x)}C-2(1+2\sin(x)).$$
Hence, the integration result is
$$\frac{u}{v}=\frac{e^{-x/2-\cos(x)}C-2(1+2\sin(x))}{e^{-x/2-\cos(x)}}=C-2e^{x/2+\cos(x)}(1+2\sin(x)),$$
as WolframAlpha yields.

• Jameson