Question

How do you evaluate the integral `1/(sqrt(49-x^2))` from 0 to `7sqrt(3/2)`?

Answer 1

Hello,

The result is `arcsin(sqrt(3/2))`, but cf. the remark at the end of my text.

Write `\frac{1}{\sqrt{49-x^2}} = \frac{1}{\sqrt{49(1-(x/7)^2)}} = \frac{1}{7\sqrt{1-(x/7)^2}}`

Change the variable in the integral :
Take `u = x/7`, therefore :
- `du = \frac{d x}{7}` or `dx = 7du`,
- if `x=0`, then `u = 0`,
- if `x=7\sqrt{3/2}`, then `u=sqrt(3/2)`.

So, `\int_0^{7sqrt(3/2)} \frac{dx}{sqrt(49-x^2)} = \int_0^{sqrt(3/2)}\frac{7 du}{7\sqrt{1-u^2}} = \int_0^{sqrt(3/2)}\frac{du}{sqrt{1-u^2}}`

You recognize the derivative of `arcsin`, so

`\int_0^{7sqrt(3/2)} \frac{dx}{sqrt(49-x^2)} = [arcsin(u)]_0^{sqrt(3/2)} = arcsin(sqrt(3/2))`.

Remark. If it's `7sqrt(3)/2` and not `7sqrt(3/2)`, the result is easier, because `arcsin(sqrt(3)/2) = pi/3`.

Answer 2

The answer is: `pi/3`.

I think there is a little mistake in your writing, are you sure that the limits of integration weren't `0` and `7sqrt3/2`?

`int_0^(7sqrt3/2)1/sqrt(49-x^2)dx=int_0^(7sqrt3/2)1/sqrt(49(1-x^2/49))dx=`

`int_0^(7sqrt3/2)1/(7sqrt(1-(x/7)^2))dx=int_0^(7sqrt3/2)(1/7)/sqrt(1-(x/7)^2)dx=`

`=[arcsin(x/7)]_0^(7sqrt3/2)=arcsin(sqrt3/2)-arcsin0=pi/3`.

This is remembering:

`int(f'(x))/sqrt(1-[f(x)]^2)dx=arcsinf(x)+c.`