Question

How do you find the integral of `int 5/sqrt(9-x^2)dx`?

Answer 1

Let `x=3*sint` hence `dx=3cost*dt` then we have that

`int 5/sqrt(9-(3sint)^2)*3cost*dt`

`int 5/sqrt(9-9*sin^2t)*3cost*dt`

`int 5/(sqrt9*sqrt(1-sin^2t))*3*cost*dt`

`int 5/(3*sqrt(cos^2t))*3*cost*dt`

`int 5/(3*cost)*3*cost*dt`

`int 5/(cancel3*cancelcost)*cancel3*cancelcost*dt`

`int 5dt=5*t+c`

But hence `sint=x/3` or `t=arcsin(x/3)`

We have that

`int 5/(sqrt(9-x^2))*dx=5*arcsin(x/3)+c`

Answer 2

Please see the explanation.

Explanation:

You can find the general form of the integral in any list of integrals:

`int1/sqrt(a^2 - x^2)dx = sin^-1(x/a) + C`

However, I think you want to be shown how to do the trigonometric substitution that put this general form on the list. Let's begin:

Given: `int5/sqrt(9 - x^2)dx = ?`

Let `x = 3sin(theta)`, then `dx = 3cos(theta)d"theta`

Substitute the above into the integral:

`5int(3cos(theta))/sqrt(9 - 9sin^2(theta))d"theta =`

Factor 9 from under the radical:

`5int(3cos(theta))/(3sqrt(1 - sin^2(theta)))d"theta =`

Use the identity `cos(theta) = sqrt(1 - sin^2(theta)`:

`5int(3cos(theta))/(3cos(theta))d"theta =`

The numerator and denominator cancel to become 1:

`5intd"theta = 5theta + C`

To reverse the substitution, we need to describe `theta` in terms of x so we start with equation that we used to make the substution:

`x = 3sin(theta)`

`x/3 = sin(theta)`

`theta = sin^-1(x/3); -3 <=x <= 3`

`int5/sqrt(9 - x^2)dx = 5sin^-1(x/3) + C; -3 <=x <= 3`