Evaluate the indefinite integral $\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx $

Question

Evaluate the indefinite integral:

$$\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx $$

(using substitution)

The answer is: $$\frac {1}{24} \cos\left(\frac{12}{x}\right) + C $$

Whereas I get a slightly different one.

Here's my solution:

$$u = \frac{6}{x}$$

$$du = - \frac{6}{x^2} \cdot dx$$

$$-\frac {1}{6} du = \frac {1}{x^2} \cdot dx$$

Making substitution:

$$ \int -\frac{1}{6} du \sin (u) \cos (u) $$

adding a new variable for substitution:

$$s = \cos (u)$$

$$ds = -sin(u) du$$

Making substitution:

$$ \frac {1}{6} \int ds \cdot s $$

Evaluating:

$$ \frac{1}{6} \cdot \frac{s^2}{2} = \frac{s^2}{12} = \frac{\cos^2(u)}{12} = \frac{\cos^2(\frac{6}{x})}{12}$$

Then using a half angle formula for $\cos^2$:

$$ \frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} = \frac{1}{24} + \frac{1}{24} \cdot \cos\left(\frac{12}{x}\right) + C$$

As you can see I have an additional $\frac{1}{24}$ in my answer... so what did I do wrong?

Answer 1

You missed the constant of integration all the way! It "absorbs" the $\frac{1}{24}$!

Spot the difference:

Evaluating:

$\displaystyle \frac{1}{6} \cdot \frac{s^2}{2} +c= \frac{s^2}{12} +c= \frac{\cos^2(u)}{12} +c= \frac{\cos^2(\frac{6}{x})}{12}+c$, where c is the constant of integration.

Then using a half angle formula for $\cos^2$:

$\displaystyle \frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} +c$ = $\frac{1}{24} \cdot \cos(\frac{12}{x}) + C$, where $C=c+\frac{1}{24}$

Answer 2

You did everything right, but you're constant of integration "absorbs" the $\frac{1}{24}$

$$\frac{1}{24}+C=C$$

The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!

If it helps, you could do something like:

$$\frac{1}{24}+C=D$$