Integrating Trigonometric functions (part 1)

Integration is topic that eludes several students. Many think that its those “you either see or don’t” topic. But its all practice and a bit of tricks. Let me touch on integrating trigonometric functions first and we shall start with sinx

Please don't try this in exams!

Please don’t try this in exams!

\int sinx dx = -cosx + c

Easy!

\int sin^{2}x dx

This requires double angle formula: sin^{2}A=\frac{1-cos2A}{2}

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

\int sin^{3}x dx

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

\int sin^{3}x dx = \int sinx(1-cos^{2}x)dx= \int sinx - sinxcos^{2}x dx

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

Notice that sinx is the derivative (f'(x)) of cosx.

So sinxcos^{2}x dx=\frac{-cos^{3}x}{3}+c.

Finally, \int sinx - sinxcos^{2}x dx = -cosx + \frac{cos^{3}x}{3}+c

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

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

So if you notice, this is essentially like an algorithm, and as the power increases the treatment is really quite similar.

Let’s look at cosx in the my next post!

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