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

2 Answers
Feb 19, 2015

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#.

Feb 19, 2015

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.#