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!

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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April 18th, 2016 04:41 PM

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Integrating Trigonometric functions (part 2)

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Integrating Trigonometric functions (part 2)

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