Integrating Trigonometric functions (part 2)

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.

Tell me what you think in the comments section!

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