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

Question

Evelyn Agard · Accepted Answer

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#.

Emily Ammerman · Answer

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.#

Cameron Alexander · Answer

undefined The integral evaluates to π.

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