Further Calculus

Derivatives of inverse trigonometric functions

To differentiate inverse trigonometric functions, use implicit differentiation. For example, to differentiate for :

This derivation has two important details to consider:

is defined for all . However, the derivative of at does not exist (the graph goes infinitely steep), which is shown by being undefined for .
When rewriting in terms of by using , we took the positive square root. This is because has a range of , and is positive over this domain, so we take the positive root.

Similar methods can be used to find all three inverse trigonometric function derivatives. All of the following are given in the formula book:

for ( is defined for ).
for ( is defined for ).
( is defined for all and so is the derivative)

Derivatives of inverse hyperbolic functions

To differentiate inverse inverse functions, again use implicit differentiation. For example, to differentiate for :

Note that here, the derivative of exists for all , but for , the derivative does not exist for as the graph becomes infinitely steep.

Similar methods can be used to find all three inverse hyperbolic function derivatives. All of the following are given in the formula book:

( is defined for all ).
for ( is defined for ).
for ( is defined for ).

Inverse trigonometric and hyperbolic functions for integration

By applying the fundamental theorem of calculus, we can reverse the results for differentiating the inverse functions to get new integration results:

These results can be generalised by making a linear substitution . All the following results are given in the formula book:

Note that the and results are not included:

is just the negative of , giving or , which can be shown on a graph as is a reflection of in .
can just be done by partial fractions, which gives the same result as the logarithmic form of .

Integrating forms with manipulation

Leading coefficients on the can be dealt with using reverse chain rule:

Or alternatively by factoring out the coefficient on :

Other integrands may require completing the square:

These integrals often require very careful applications of the reverse chain rule.

Integrating using partial fractions

When there is a quadratic factor in the denominator of a rational expression, there are three different possibilities:

The quadratic factors into two different linear factors, , giving partial fractions of .
The quadratic is a perfect square, , giving partial fractions of .
The quadratic does not factorise (it is irreducible), giving a fraction of the form . This can be integrated using , but could require completing the square first.

Derivations

The derivations of the 4 given results could be tested. Below are the full derivations. All of the derivations follow the same basic pattern:

Substitute
Apply or
Simplify and integrate
Substitute back for to get the desired answer.

Inverse sine

Working towards

Apply the substitution with
The substitution gives
Cancel the common factor of to get
Apply to get
Integrate to get
Substitute with to get

Inverse tangent

Working towards

Apply the substitution with
The substitution gives
Cancel the common factor of to get
Apply to get
Integrate to get
Substitute with to get

Inverse hyperbolic sine

Working towards

Apply the substitution with
The substitution gives
Cancel the common factor of to get
Apply to get
Integrate to get
Substitute with to get

Inverse hyperbolic cosine

Working towards

Apply the substitution with
The substitution gives
Cancel the common factor of to get
Apply to get
Integrate to get
Substitute with to get