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- Jan 26, 2012

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$$\int[(1-2\cos(x))(3+2\cos(x))e^{\frac{1}{2}\,x+\cos(x)}]\,dx.$$

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.