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


\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!

One Comment

Leave a Reply