The derivatives and integrals of **trigonometric functions** are covered in Maths T. So in this section, I’ll only teach you how to differentiate **inverse trigonometric functions**. A warning here is that you must study the chapter **Integration** (especially the part on **integration by parts**) in Maths T before you come to this section, if not you will get really confused.

To find the derivative of **sin ^{-1} x**, we need to make use of our knowledge on differentiating a function

**implicitly**. We let

**x = sin y**. Differentiating the function implicitly, we have

From here, you can further deduce that the derivations of the derivatives of inverse trigonometric functions should follow the same rule, i.e., differentiating the functions implicitly, then making use of their trigonometric identities. The list of derivatives of all the inverse trigonometric functions are as follows:

where **a **is a constant. You should try to prove each and every one of them as an exercise.

You should further try to differentiate these functions with complicated variables using all the differentiation rules you learnt. For example,

Take note that once you differentiate an inverse trigonometric function, it becomes a fraction of polynomials. Do not worry about the anti-derivatives of these inverse polynomial functions now, as I will give you a summary table in the section on **Reduction Formulae**.

However, I want to discuss on the anti-derivative of the inverse trigonometric function itself. For example, I want to find

To do this, you need to make use of **integration by parts.** If you followed the formula in the Maths T formula sheet, it would be

However, I suggest that you use this formula which makes you remember easier:

Before I continue, let me explain this formula. Normally, you only use integration by parts when you are trying to integrate a product of 2 functions, which are most likely **logarithmic, exponential, polynomial **and** trigonometric functions.** So in any case, you let one function be **u, **and the other function be **v**. Notice that **v** has to be a function that is easy to integrate, while **u **has to be the other one which is hard to integrate / easy to differentiate. In words, this formula can be read as

“Integration of **u × v** = [ **u × **integrate **v** ] – integration of (differentiate **u × **integrate **v**)”

Never mind if you don’t get it, as long you have your own version of I by P. So continuing on integrating **sin ^{-1} x**, we let

**u**=

**sin**, and

^{-1}x**v**=

**1**. We have

Get it? So the important tips to this question is to put **v = 1 **(you might recall that this is the method you use to integrate **ln x**). So the rest of the functions, after integration gives

Try to derive all of them as an exercise. Note that the term **ln [x + √( x ^{2} – 1)]** is actually a

**cosh**function.

^{-1}x

Do lots of practices on this. There are a lot of exam questions on integrations and differentiations, which combines various functions together. You might not want to make mistakes and lose marks. ☺

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