How do I integrate $\sin\theta + \sin\theta \tan^2\theta$? How do I integrate $\sin\theta + \sin\theta \tan^2\theta$ ?
First thing, 
I have been studying maths for business for approximately 3 months now. Since then, I studied algebra and then I started studying calculus. Yet, my friend stopped me there, and asked me to study Fourier series as we'll need it for our incoming projects. So I feel that I am missing a lot of things as I haven't studied integrals yet. 
Today, I encountered this solution that I couldn't understand at all. 
Apparently, $\int(\sin\theta + \sin\theta \tan^2\theta)d\theta = \int(\sin\theta (1 + \tan^2\theta))d\theta$. 
Below, a link to the solution at 5:08.
https://youtu.be/aw_VM_ZDeIo
He stated the we have to learn the integral identities. So, I started searching the whole internet looking for them. But, I think I couldn't find them. The only thing that I found was something called Magic hexagon. I thought of reading about $\theta$ as it might mean something. But, after all I learned that it is just a regular greek letter used as a variable.
 A: What happens in the video is that the $\sin(x)$ is factored out, and so it becomes $\sin(x)(1+\tan^2(x))$. This is simply factoring, and by expanding you can see it is equivalent to $\sin(x)+\sin(x) \tan^2(x)$. 
As you said, the Greek letter $\theta$ is just a common variable, which is often used in trigonometric problems instead of the standard “$x$”. Many integrals of this kind require use of trigonometric identities, and if you wish to review them a list can be found here. 
A: $$
\sin\theta+\sin\theta\tan^2\theta=\sin\theta(1+\tan^2\theta)=\sin\theta\sec^2\theta=\sec\theta\tan\theta
$$
because $(1+\tan^2\theta)=\sec^2\theta$ is a 'standard' identity.
Also
$$
\frac{d\sec\theta}{d\theta}=\sec\theta\tan\theta
$$
is a 'standard' derivative.
Therefore
$$
\int(\sin\theta+\sin\theta\tan^2\theta)\, d\theta = \int \sec\theta\tan\theta\, d\theta=\sec\theta+C
$$
where $C$ is an arbitrary constant
A: Bioche's rules  say we can make the substitution $u=\cos\theta$, $\mathrm d u=-\sin\theta\,\mathrm d\theta$. Indeed
$$ \int(1+\tan^2\theta)\sin\theta\,\mathrm d\theta=\int \frac1{\cos^2\theta}\,\sin\theta\,\mathrm d\theta=\int -\frac{\mathrm du}{u^2}=\frac1u=\frac1{\cos\theta}. $$
A: This is a step-by-step solution using only basic facts.

$$\tan(\theta) = \frac{\sin\theta}{\cos\theta}$$

Then:
$$\int\sin\theta (1 + \tan^2\theta)d\theta = \int\sin\theta \left(1 + \frac{\sin^2\theta}{\cos^2\theta}\right)d\theta = \int\sin\theta \left(\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)d\theta.$$

$$\sin^2\theta + \cos^2\theta = 1$$

Then:
$$\int\sin\theta (1 + \tan^2\theta)d\theta = \int\sin\theta\left(\frac{1}{\cos^2\theta}\right)d\theta = \int\sin\theta\cos^{-2}\theta d\theta.$$

$$\int f'(x) f^n(x) dx = \frac{f^{n+1}}{n+1} + c.$$

Then:
$$\int\sin\theta (1 + \tan^2\theta)d\theta = \frac{1}{\cos\theta}+c.$$
