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.