# 6. Integration: Inverse Trigonometric Forms

by M. Bourne

Using our knowledge of the derivatives of inverse trigonometric identities that we learned earlier and by reversing those differentiation processes, we can obtain the following integrals, where u is a function of x, that is, u=f(x).

int(du)/sqrt(a^2-u^2)=sin^(-1)u/a+K

int(du)/(a^2+u^2)=1/atan^(-1)u/a+K

NOTE: Your calculator has sin^(-1) and tan^(-1) buttons, but these create quite a bit of confusion because they are inverse functions, not reciprocals. We could also (better) write these formulas using arcsin and arctan as follows:

int(du)/sqrt(a^2-u^2)=arcsin\ u/a+K

int(du)/(a^2+u^2)=1/a arctan\ u/a+K

### Example 1

Integrate: int(dx)/sqrt(49-x^2

### Example 2

Integrate: int_0^1(2\ dx)/(sqrt(9-4x^2

### Example 3

Find the area bounded by the curve y=1/(1+x^2) and the lines x = 0, y = 0 and x = 2.

### Caution

There are a number of integrals of forms which look very similar to the above formulas but are actually different, e.g.

int(dx)/(sqrt(x^2-1)),\ \ int(dx)/(sqrt(1+x^2)),\ \ int(dx)/(1-x^2),\ "etc."

We will develop methods to solve these in a later section. (See Integration by Trigonometric Substitution.)

### Exercises

Integrate each of the given functions:

1. int(3\ dx)/(25+16x^2)

2. int(2\ dx)/(x^2+8x+17)

3. int(dx)/(sqrt(2x-x^2)

4. Find the area bounded by the curve ysqrt(4-x^2)=1 and the lines x = 0, y = 0 and x = 1.

This is not the answer ... it is a corruption of a .gif file containing the answer!!:

[This happened a long time ago in MS Word during a period of great instability - the computer's and mine...^_^]

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