### 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!

### Importance of Prelims

Many students have the mentality that now they are in JC2, what matters is just how you do in A-levels since its what you get into University. I thought I should share some pros of doing well in Prelims.

Firstly, doing well in prelims allow students to obtain early admission into universities. I know a handful of students getting placements prior to receiving results.

Secondly, students that are aiming for scholarships, your teachers’ testimonials are very crucial to your application. And doing well in prelims, usually allows you to receive good testimonials from them.

Lastly, who doesn’t want some confidence booster going to A-levels!

If you’re having problems getting ready for A-levels, consider getting help soon! The 20H Crash Course information can be found here!

### Easiest way to remember cosec, sec and cot

Here is a little tip to how to remember the formulas for your favorite trigonometry functions!

$\frac {1}{sinx} = cosecx$

$\frac {1}{cosx} = secx$

$\frac {1}{tanx} = cotx$

### MF15 can’t be more useful!

The only cheatsheet every student can bring into the exam hall.

### What is MF15?

It’s the formula sheet every H2 Math student needs to know about.

View PDF

I thought we can look at the MF15 and also highlight a few common mistakes that students make in exams. 🙂

Firstly, the trigonometry formulas are all given in exams. Most students however, do not know what they have. I strongly suggest students spend time to know what they have. And if you’re interested, you should really learn how to derive the sum to product formula from the product to sum formula. You can read more here.

Secondly, the integration formulas that students see in MF15, are not as innocent as they look. Many students overlook the fact that the coefficient of the $x$ is ONE! You can read up more here. too.

Lastly, the r-formula is given. Students should know how to use it at the least. Also the properties of the Binomial and Poisson distribution are given (if you’re J2 and still need to refer to know the E(X) and Var(X), you should be worried too).

And if you’re not using TI-84, I strongly suggest you learn how to read the T-Table and Z-Table.

### 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!

### How to derive the sum to product formula

We shall look at an easy method of remembering or knowing how to derive the sum to product formula. As you see below is the product to sum formula:

We first let
$\alpha = {\frac{1}{2}({P+Q})}$
$\beta = {\frac{1}{2}({P-Q})}$, and thus have,
$\alpha + \beta = P$
$\alpha - \beta = Q$

The first formula should look like this:
$sin(\alpha+\beta) + sin(\alpha-\beta)=2{\mathrm{sin}\alpha}{\mathrm{cos}\beta}$

$\frac{1}{2}$ is being taken care of too but be careful that there is a “2” there.

This little switcheroo will allow us to be at the doorsteps of the sum to product formula. Students should attempt the substitution yourself, and you will realise that it is not that hard after all. This should ease your life of finding P’s and Q’s. Try deriving the remaining three formulas!

Please comment at the bottom if you run into any questions, I will love to help! 🙂