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)`.
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)/(a^2+u^2)=1/a arctan\ u/a+K`
Continues below ⇩
This is the graph of the function we just integrated.
Graph of `y(x)=1/sqrt(49-x^2)`.
The next graph is a typical solution graph for the integral we just found, with `K=0`.
Graph of `y(x)=arcsin(x/7)`.
Integrate: `int_0^1(2\ dx)/(sqrt(9-4x^2`
Here is the graph of the integral we just found. It represents the area under the curve `y(x)=2/sqrt(9-4x^2)` from `0 < x < 1`.
Graph of `y(x)=2/sqrt(9-4x^2)`.
Find the area bounded by the curve `y=1/(1+x^2)` and the lines x = 0, y = 0 and x = 2.
This is the area we found just now.
Graph of `y(x)=1/(1+x^2)` showing area between `0 < x < 2`.
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.)
Integrate each of the given functions:
1. `int(3\ dx)/(25+16x^2)`
2. `int(2\ dx)/(x^2+8x+17)`
4. Find the area bounded by the curve `ysqrt(4-x^2)=1` and the lines `x = 0`, `y = 0` and `x = 1`.