# Trig integral $\int ( \cos{x} + \sin{x}\cos{x}) \, dx$

Assume we have:

$$\int (\cos{x} + \sin{x}\cos{x}) \, dx$$

Two ways to do it:

Use $$\sin{x}\cos{x} = \frac{ \sin{2x} }{2}$$ Then $$\int \left(\cos{x} + \frac{\sin{2x}}{2} \right) \, dx = \sin{x} - \frac{ \cos{2x} }{ 4 } + C.$$ The other way, just see that $$u = \sin(x), du = \cos(x)dx$$, $$\int ( \cos{x} + \sin{x}\cos{x}) \, dx = \sin{x} + \frac{\sin^2{x}}{2} + C.$$

Now the part I don't see fully is, why aren't these results completely equal?

Taking the 2nd result, \begin{align} \sin{x} + \frac{\sin^2{x}}{2} &= \sin{x} + \frac{1}{2} \, \left( \frac{1 - \cos{2x}}{2} \right) \\ &= \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4}. \end{align}

So you have to absorb $$\frac{1}{4}$$ into $$C$$ for them to be equal.

Shouldn't they be equal straight away?

• Why should the two $C$'s be equal? (Ceci n'est pas cette 'c'!) When dealing with indefinite integrals, what's important is that you end up with two expressions with identical derivatives, which in this case they are. Apr 15, 2011 at 18:56

Your question boils down to this:

If $\int f(x) dx = F(x) + C$ and $\int f(x) dx = G(x) +C$ are both correct, then shouldn't it be true that $F(x)=G(x)$?

The answer is no. $\int f(x)dx=F(x) +C$ means that (on the relevant interval, in this case all of $\mathbb{R}$) every antiderivative of $f(x)$ has the form $F(x)+C$ for some constant $C$. The indefinite integral is really referring to a set of functions, namely all of the functions (on the relevant interval) whose derivatives equal $f(x)$. If $F'(x)=f(x)$, then that set can be written as $\{F(x)+C:C\in \mathbb{R}\}$. But the set of functions of the form $F(x)+C$ for some constant $C$ is precisely the same as the set of functions of the form $F(x)+22+C$ for some constant $C$, for instance. Explicitly, $\{F(x)+C:C\in\mathbb{R}\}=\{F(x)+22+C:C\in\mathbb{R}\}$. That is, if $F(x)$ and $G(x)$ only differ by a constant and $F(x)$ is an antiderivative of $f(x)$, then $\int f(x)dx=G(x)+C$ is also correct.

• Ran into an even more illustrative one: $\int{ \frac{ -2x }{ (1+x^2)^2 } } dx$ becomes either $\frac { -x^2 }{ 1 + x^2 }$ or $\frac{ 1 }{ 1 + x^2 }$. Both work. Apr 21, 2011 at 22:53
• @bobobobo: Good example. If you have $\frac{f(x)}{g(x)}$ where $f$ and $g$ are polynomials of the same degree, then polynomial division gives you $k+\frac{r(x)}{g(x)}$ where $k$ is constant and $r$ has lower degree than $g$. Thus $\frac{r(x)}{g(x)}$ and $\frac{f(x)}{g(x)}$ have the same derivative. Apr 22, 2011 at 15:36

If you write

$\int{ \cos{x} + \frac{\sin{2x}}{2} dx }=\sin{x} - \frac{ cos{2x} }{ 4 } + C_1$

and

$\int{ \cos{x} + \sin{x}\cos{x} dx }= \sin{x} + \frac{\sin^2{x}}{2} + C_2$,

then you can see what's going on with the constants in your two cases($C_1$ and $C_2$). Being careful with choosing the notation is good for understanding.