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

Hint: You must use integration by parts.

Cultivating Champions, Moulding Success

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

Hint: You must use integration by parts.

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

or .

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,

.

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.

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

(MF15)

(MF15)

So how do we ? I’ll first rewrite it as for some insights.

We can’t adopt the method here. So, Integration by parts?

Wait! again? hmmm.

So we have that

.

Then with a bit of juggling and manipulations, we have

.

I do hope this gives you some insights. You should try on your own using the information here.

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

(MF15)

(MF15)

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

!

Here we have a slight problem with

Recall the method we saw in part 1 and 2…

So,

As you probably guessed, and for , we can refer to the previous results.

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

As promised, we will look at this time. It might get boring, as the method is exactly the same as as they are quite related.

Easy!

This requires double angle formula:

Here we introduce trigo identity:

Here we have a problem!

But recall we did some really similar in part 1, and notice that is the derivative () of .

So .

Finally,

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 and .

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 method to integrate. Just saying, .

Tell me what you think in the comments section!

Many students often overlook that the coefficient of 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. š

For my careless students, I usually recommend they make the case of the coefficient of be ONE instead. So and after applying formula gives, which will give the same answers after simplifications.

You may practice a few of the following questions!

Let me know if you have problems!

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

Easy!

This requires double angle formula:

Here we introduce trigo identity:

Here we have a problem!

Notice that is the derivative () of .

So .

Finally,

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 in the my next post!