Can we simplify this integral using trig substitution and partial fractions?

[courtrigrad]

\int \frac{dx}{x\sqrt{a^{2}+x^{2}}}.

So, x = a\tan\theta, and dx = a\sec^{2}\thetha d\theta. When we substitute we get: \int\frac{a\sec^{2}\theta}{(a\tan\theta)(a\sec\theta}) which equals \frac{1}{a}\int \csc \theta d\theta. I know that \int \csc \theta d\theta = -\ln|\csc\theta + \cot\theta|. And \theta = \tan^{-1}(\frac{x}{a}). So I substitute this into the equation. How do we get from that to this:

(\frac{1}{a})\ln|\frac{x}{a+\sqrt{a^{2}+x^{2}}}

Thanks

 

[StatusX]

Can you write cot(\theta) and csc(\theta) in terms of tan(\theta)?

 

[courtrigrad]

yeah and I got \frac{\cos\theta}{\sin\theta}+\frac{1}{\sin\theta} = \frac{1+\cos\theta}{\sin\theta}

 

[StatusX]

Not in terms of \theta, in terms of tan(\theta). For example, cot(\theta)=1/tan(\theta). What is csc(\theta)? (hint: use sec²(\theta)=1+tan²(\theta) )

 

[courtrigrad]

-ln(\frac{1}({\tan\theta}+\sqrt{1+(\frac{1}{\tan\theta})^{2})

 

[courtrigrad]

i got it. thanks

 

[benorin]

x=a\tan \theta gives \tan \theta = \frac{x}{a} so we have a right triangle where: the length of the leg oppsite \theta is x, the length of the leg adjacent to \theta is a, and the hypotenuse is \sqrt{a^2+x^2}. From the triangle it follows that \cot \theta = \frac{a}{x}, what is \csc \theta?

 

[0rthodontist]

An alternative way of solving it is to let
u = \sqrt{x^2 + a^2}
du = \frac{x}{\sqrt{x^2 + a^2}}dx
Then the integral becomes
\int \frac{du}{u^2-a^2}
which can be solved by partial fraction decomposition.