# Math Help - How to integrate this?

1. ## How to integrate this?

Question is in orange, answer is in black.

I got no idea how they got this answer :\

The way im trying is using a substition of $kx^2 = gsin\theta$

Just the one question i cant get my head aroung in this exercise step by step method would be appreciated

2. Let $x=\sqrt{\frac{g}{k}}sec{\theta}, \;\ dx=\sqrt{\frac{g}{k}}sec{\theta}tan{\theta}d{\thet a}$

and it should whittle down nicely.

3. Originally Posted by Kevlar
Question is in orange, answer is in black.

I got no idea how they got this answer :\

The way im trying is using a substition of $kx^2 = gsin\theta$

Just the one question i cant get my head aroung in this exercise step by step method would be appreciated
first lets pull some stuff out to get

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

For simplicity 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 we get

$\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=$

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

So finally we get

$\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)$

4. Originally Posted by galactus
Let $x=\sqrt{\frac{g}{k}}sec{\theta}, \;\ dx=\sqrt{\frac{g}{k}}sec{\theta}tan{\theta}d{\thet a}$

and it should whittle down nicely.
still got no idea this just completly confused me

5. Originally Posted by TheEmptySet
first lets pull some stuff out to get

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

For simplicity 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 we get

$\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=$

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

So finally we get

$\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)$
How do you get from the ln to inverse tanh?

6. Originally Posted by Kevlar
still got no idea this just completly confused me
That's just trig substitution. EmptySet has a nicer solution.

Note the identity $tanh^{-1}(x)=\frac{1}{2}ln(\frac{1+x}{1-x})$

But, note similarity with $\int\frac{1}{a^{2}+u^{2}}du=\frac{1}{a}tan^{-1}(\frac{u}{a})+C$