MHB Week #140: How to Solve This Challenging Integral?

[Ackbach]

Here is this week's problem.

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

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

MarkFL answered this week's POTW correctly. You can see his solution below:

Spoiler

We are given to evaluate:

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

Let's rewrite the integrand. First, find the product of the first two factors:

$$\left(3-4\cos(x)-4\cos^2(x)\right)e^{\frac{x}{2}+\cos(x)}$$

Apply a Pythagorean identity:

$$\left(3-4\cos(x)-4\left(1-\sin^2(x)\right)\right)e^{\frac{x}{2}+\cos(x)}$$

$$\left(3-4\cos(x)-4+4\sin^2(x)\right)e^{\frac{x}{2}+\cos(x)}$$

$$\left(4\sin^2(x)-1-4\cos(x)\right)e^{\frac{x}{2}+\cos(x)}$$

Factor out a $-2$ and within the parentheses, add zero in the form of $\sin(x)-\sin(x)$:

$$-2e^{\frac{x}{2}+\cos(x)}\left(\sin(x)-2\sin^2(x)+\frac{1}{2}-\sin(x)+2\cos(x)\right)$$

Factor:

$$-2e^{\frac{x}{2}+\cos(x)}\left((2\sin(x)+1)\left(\frac{1}{2}-\sin(x)\right)+2\cos(x)\right)$$

Distribute the exponential:

$$-2\left((2\sin(x)+1)e^{\frac{x}{2}+\cos(x)}\left(\frac{1}{2}-\sin(x)\right)+2\cos(x)e^{\frac{x}{2}+\cos(x)}\right)$$

Write as the differentiation of a product:

$$\frac{d}{dx}\left(-2(2\sin(x)+1)e^{\frac{x}{2}+\cos(x)}\right)$$

And so , we may state:

$$\int\left[(1-2\cos(x))(3+2\cos(x))e^{\frac{x}{2}+\cos(x)}\right]\,dx=\int\,d\left(-2(2\sin(x)+1))e^{\frac{x}{2}+\cos(x)}\right)=-2(2\sin(x)+1)e^{\frac{x}{2}+\cos(x)}+C$$

Here is an alternative solution from yours truly:

Spoiler

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.