### Integration Question #1

This is a rather easy questions which stumped a good number of students initially.

$\int x^{n}lnx dx$

Hint: You must use integration by parts.

### Integrating Trigonometric functions (part 5)

This will be about integrating product of trigonometric functions with no powers involved for example,

$\int sin6xcos21x dx$ or $\int cosxcos5x dx$.

The simple trick here is to use the product to sum to formulas that can be found here.
For convenience, I’ll insert them here.

So very conveniently,

$\int cosxcos5x dx=\int \frac{1}{2}(cos4x+cosx) dx = \frac{1}{2}(\frac{sin4x}{4}+sinx)$.
Now you see the usefulness of the product to sum formula! Better learn how to find them!

For trigonometric functions with powers involved, you can refer to part 1 to part 5.

### Integrating Trigonometric functions (part 4)

We shall now proceed to integrating $secx$ and similarly, lets refresh the formulas we should know.

$\frac {d}{dx}tanx = sec^{2}x$

$\frac {d}{dx}secx = secxtanx$

$\int tanx dx = ln|secx|+c$ (MF15)

$\int secx dx = ln|secx+tanx|+c$ (MF15)

$\int sec^{2}x dx = tanx+c$

$\int sec^{3}x dx = \int secx(sec^{2}x)dx = \int secx(tan^{2}x+1)dx = \int secxtan^{2}x+secx dx$
So how do we $\int secxtan^{2}x dx$? I’ll first rewrite it as $\int (secxtanx)(tanx)dx$ for some insights.

We can’t adopt the $\int f'(x)f(x) dx$ method here. So, Integration by parts?

$\int (secxtanx)(tanx)~dx$

$= secx(tanx) - \int secx(sec^{2})~ dx$

$= secxtanx-\int sec^{3}x~dx$

Wait! $\int sec^{3}x dx$ again? hmmm.

So we have that

$\int sec^{3}x ~dx$

$= \int secxtan^{2}x+secx ~dx$

$= secxtanx-\int sec^{3}xdx + \int secx ~dx$.

Then with a bit of juggling and manipulations, we have

$2\int sec^{3}x dx = secxtanx + ln|secx+tanx|+c$.

I do hope this gives you some insights. You should try $\int sec^{4}x dx$ on your own using the information here.

### Integrating Trigonometric functions (part 3)

So part 3, we look at $tanx$! This is slightly more complicated as we need to first lay out some formulas that we should know.

$\frac {d}{dx}tanx = sec^{2}x$

$\frac {d}{dx}secx = secxtanx$

$\int tanx dx = ln|secx|+c$ (MF15)

$\int secx dx = ln|secx+tanx|+c$ (MF15)

Notice how $tanx$ and $secx$ are quite related here… And we recall we have a trigonometry identity that happens to be that closely related:

$tan^{2}x+1=sec^{2}x$!

$\int tan^{2}x dx = \int sec^{2}x -1 ~dx = tanx - x + c$

$\int tan^{3}x ~dx$

$= \int tanx(tan^{2}x) ~dx$

$= \int tanx(sec^{2}x-1) ~dx$

$= \int tanxsec^{2}x-tanx ~dx$

Here we have a slight problem with $\int tanxsec^{2}x$

Recall the $\int f'(x)f(x)$ method we saw in part 1 and 2…
$\int tanxsec^{2}x= \frac {tan^{2}x}{2}+c$

So, $\int tanxsec^{2}x-tanx ~dx=\frac {tan^{2}x}{2}-ln|secx|+c$

$\int tan^{4}x dx = \int tan^{2}x(sec^{2}x-1) ~dx = \int tan^{2}xsec^{2}x-tan^{2}x dx$

As you probably guessed, $\int tan^{2}xsec^{2}x ~dx = \frac {tan^{3}x}{3}+c$ and for $\int tan^{2}x dx$, we can refer to the previous results.

We will look at $secx$ in part 4, and surprise surprise, it will be quite a different appraoch! Let me know if there are any questions.

### 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!

### Let’s try some calculus questions

Many students often overlook that the coefficient of $x$ in the integration or differentiation formulas in MF15 is 1!!! When it is not 1, many things changes. I’ll let the examples do the talking. 🙂

### Differentiation (recall chain rule)

$\frac {d}{dx}(sin^{-1}(3x^2)) = \frac {1}{\sqrt{1-(3x^2)^2}}(6x)$

### Integration

$\int \frac {1}{4+9x^2} dx = \frac {1}{2} {tan^{-1}(\frac {3x}{2})}\times\frac{1}{3}$

For my careless students, I usually recommend they make the case of the coefficient of $x$ be ONE instead. So $\int \frac {1}{4+9x^2} dx = \frac{1}{9}\int \frac {1}{\frac{4}{9}+x^2} dx$ and after applying formula gives, $(\frac{1}{9})(\frac{1}{\frac{2}{3}}){tan^{-1}(\frac {x}{\frac{2}{3}})}$ which will give the same answers after simplifications.

### Practice

You may practice a few of the following questions!

$\int\frac {1}{4-9x^2}dx$

$\int\frac {1}{9x^{2}-4}dx$

$\int\frac {1}{\sqrt{4-9x^2}}dx$

$\int\frac {1}{2x^{2}-2x-10}dx$

Let me know if you have problems!

### 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 $sinx$

$\int sinx dx = -cosx + c$

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!