As promised, we will look at $cosx$ this time. It might get boring, as the method is exactly the same as $sinx$ as they are quite related.

$\int cosx dx = sinx + c$

Easy!

$\int cos^{2}x dx$

This requires double angle formula: $cos^{2}A=\frac{1+cos2A}{2}$

$\int cos^{2}x dx = \int\frac{1+cos2x}{2}dx = \frac{1}{2}(x+\frac{sin2x}{2})+c$

$\int cos^{3}x dx$

Here we introduce trigo identity: $sin^{2}x + cos^{2}x = 1$

$\int cos^{3}x dx = \int cosx(1-sin^{2}x)dx= \int cosx - cosxsin^{2}x dx$

Here we have a problem! $cosxsin^{2}x dx=?$

But recall we did some really similar in part 1, and notice that $cosx$ is the derivative ($f'(x)$) of $sinx$.

So $cosxsin^{2}x dx=\frac{sin^{3}x}{3}+c$.

Finally, $\int cosx - cosxsin^{2}x dx = sinx - \frac{sin^{3}x}{3}+c$

$\int cos^{4}x dx = \int (cos^{2}x)(cos^{2}x) dx$

Here we can apply double angle a few times to break it down before integrating.

After seeing both part 1 and part 2, you should notice some intuitive method.

Consider $\int sin^{n}x dx$ and $\int cos^{n}x dx$.

Should n be even, we introduce the double angle formula to simplify things.
Should n be odd, we introduce the trigonometry identities and integrate. We must apply ${f'(x)}{f(x)}$ method to integrate. Just saying, $\int sinx cos^{17}x dx = \frac{-cos^{18}x}{18}+c$.

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