Need help on integration question i found on net

  • Thread starter Thread starter Keval
  • Start date Start date
  • Tags Tags
    Integration Net
Keval
Messages
22
Reaction score
0
Question is in orange, answer is in black.
2hhen4i.png

I got no idea how they got this answer :\

The way I am trying is using a substition of kx^2 = gsin\theta

Just the one question i can't get my head aroung in this exercise step by step method would be appreciated
 
Physics news on Phys.org
Hi Keval! :smile:

Not sin, but tanh … try substituting x√(k/g) = tanhu :wink:
 
With denominator instead k x^2 + g you get an arctan answer, right? Use the same steps on this one and get an atanh answer.
 
someone check this out for me ??

-\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}


to simplify i let\frac{g}{k}=m^2 to get


-\frac{1}{k}\int \frac{dx}{x^2-m^2}=-\frac{1}{k}\int \frac{dx}{(x-m)(x+m)}


Using partial fractions i got
\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=



\frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=

-\frac{1}{m}\cdot \frac{1}{2}\ln \left|\frac{x+m}{x-m} \right|=-\frac{1}{m}\tanh^{-1}\left( \frac{x}{m}\right)

\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)
 
Hi Keval! :wink:

Yes, your partial fractions method is fine. :smile:

But it would be far quicker to start "let x√(k/g) = tanhu, dx√(k/g) = sech2u du" …

try it! :smile:
 
Back
Top