# 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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