How do I integrate this equation with respect to $dcos\theta$? I have this equation:
$$\int_0^{-1}\big[(1+\cos^2\theta)+\cos\theta\big]d\cos\theta$$
I tried to do a u substitution. However, this didn't work. How can I evaluate this integral?
Edit: Also attempted to use Wolfram Alpha. It didn't recognize that I wanted to use $d\cos\theta$ and not $d\theta$.
 A: It seems to be a Riemann–Stieltjes integral. Because $\cos(\theta)$ is continuously differentiable, we have that
$$\int_0^{-1}\left[(1+\cos^2\theta)+\cos\theta\right]\mathrm{d}\cos\theta=\int_0^{-1}\left[(1+\cos^2\theta)+\cos\theta\right]\frac{\mathrm{d}\cos\theta}{\mathrm{d}\theta}\mathrm{d}\theta$$
A: As has been pointed out by @Botond and reinforced by @zwim , this is an example of a Riemann-Stieltjes integral which takes the form 
$$\int_a^bf(\theta) \ dg(\theta)$$
In this case, the variable is $\theta$, $f(\theta)=1+ \cos^2(\theta)+\cos\theta$, and $g(\theta)=\cos\theta$
As Botond pointed out,
$$\int_a^bf(\theta) \ dg(\theta) = \int_a^bf(\theta)g'(\theta) \ d\theta$$
Thus, for your specific example, we have 
$$\begin{align}\int_0^{-1}[1+ \cos^2(\theta)+\cos\theta] \ d\cos(\theta)&=-\int_0^{-1}\sin(\theta)[1+ \cos^2(\theta)+\cos(\theta)] \ d\theta
\\
&=-\int_0^{-1}[\sin(\theta) + \sin(\theta) \cos^2(\theta)+\sin(\theta)\cos(\theta)] \ d\theta
\\
&=\cos(\theta)\Bigg|_0^{-1} + \dfrac{1}{3}\cos^3(\theta)\Bigg|_0^{-1}+ \dfrac{1}{2}\cos^2(\theta)\Bigg|_0^{-1}
\\
&=\cos(-1) + \dfrac{1}{3}\cos^3(-1)+ \dfrac{1}{2}\cos^2(-1) - 1 - \dfrac{1}{3} - \dfrac{1}{2}
\\
&=\cos(-1) + \dfrac{1}{3}\cos^3(-1)+ \dfrac{1}{2}\cos^2(-1) - \dfrac{11}{6}
\end{align}$$
