Question

How do you integrate `int 1/sqrt(25-t^2)` by trigonometric substitution?

Answer 1

`int (dt)/sqrt(25-t^2) = arcsin(t/5) +C`

Explanation:

Substitute:

`t=5sinx`
`dt = 5cosx`

so that:

`int (dt)/sqrt(25-t^2) = int (5cosxdx)/sqrt(25-25sin^2x)`

`int (dt)/sqrt(25-t^2) = int (5cosxdx)/sqrt(25(1-sin^2x))`

`int (dt)/sqrt(25-t^2) = int (cancel5cosxdx)/(cancel5sqrt(cos^2x))`

Now, the integrand function is defined only for `t in (-5,5)`, so that we can assume `x in (-pi/2,pi/2)`. In this interval `cosx` is always positive, so:

`sqrt(cos^2x) = cosx`

`int (dt)/sqrt(25-t^2) = int (cosxdx)/(cosx) = int dx = x+C`

Undoing the substitution:

`int (dt)/sqrt(25-t^2) = arcsin(t/5) +C`