This subsection is not completed yet, but in the meantime here are some practice questions.
Remember:
\begin{align*}
\int \sin x \dif x &= -\cos x + c\\
\int \cos x \dif x &= \sin x + c\\
\int \sec^2 x \dif x &= \tan x + c\\
\end{align*}
My line of thinking when I remember how to integrate sine goes like this: it’s something to do with $\cos x$, but the derivative of $\cos x$ is $-\sin x$, so I have to make it negative to compensate for the minus, so it’s $-\cos x$.
Remember:
When integrating $\sin^2 x$ or $\cos^2x$, write it in terms of $\cos2x$ instead (using the double-angle formulae from Section 11.4.2: Angle Formulae).
\begin{align*}
\sin^2x &= \frac{1-\cos2x}{2}\\
\cos^2x &= \frac{1+\cos2x}{2}\\
\end{align*}
You can memorise these formulae, or you can derive them from $\cos2x = 2\cos^2x - 1 = 1-2\sin^2x$ each time.