[SPLIT] int [ 1 / (x^4 sqrt[x^2 + 3]) ] dx

[warwick]

\(\displaystyle \L\\\int{(\frac{dx}{x^{4}sqrt{x^{2}+3}}})\)

\(\displaystyle \L\\x=sqrt{3}tan{\theta}\)

\(\displaystyle \L\\{\sqrt{3}sec^{2}{\theta}d{\theta}\)

\(\displaystyle \L\\{\theta}=tan^{-1}(\frac{x}{\sqrt{3}})\)

\(\displaystyle \L\\\\\int{(\frac{sqrt{3}sec^{2}{\theta}d{\theta}}{{9}{tan^{4}{\theta}sqrt{3sec^{2}{\theta}}})\)

\(\displaystyle \L\\\\\frac{1}{9}\int{(\frac{sec{\theta}}{tan^{4}{\theta}}})d{\theta}\)

 

[galactus]

Hey again Warwick:

You are good so far. Havin' trouble with the wacky trig what-nots?.

\(\displaystyle \L\\\frac{1}{9}\int\frac{sec(x)}{tan^{4}(x)}dx\\=\frac{1}{9}\int[sec(x)cot^{4}(x)]dx\\=\frac{1}{9}\int[cot^{3}(x)csc(x)]dx\\=\frac{1}{9}\int[csc^{3}(x)cot(x)]dx-\frac{1}{9}\int[cot(x)csc(x)]dx\)

Now, let \(\displaystyle \L\\u=csc(x), \;\ du=-csc(x)cot(x)dx\)

Then you have a nice little easy integral:

\(\displaystyle \L\\\frac{1}{9}[-\int{u^{2}}du+\int{du}]\)

\(\displaystyle \L\\\frac{1}{9}[\frac{-1}{3}u^{3}+u]\)

\(\displaystyle \L\\\frac{-1}{27}csc^{3}(x)+\frac{1}{9}csc(x)\)

 

[warwick]

Then you have a nice little easy integral:...

\(\displaystyle \L\\\frac{-1}{27}csc^{3}(x)+\frac{1}{9}csc(x)\)

Shouldn't it be cot^3 not csc^3? I got as far as cot^3 and csc originally. I must be missing some sort of identity.

 

[galactus]

No, I don't think. It's csc^3(x)

Plug in your originals to get back to the original terms. I used x here instead of theta for ease of typing.

If we sub \(\displaystyle \L\\tan^{-1}(\frac{x}{\sqrt{3}})={\theta}\) bacl in we get:

\(\displaystyle \L\\\frac{\sqrt{x^{2}+3}(2x+3)}{27x^{3}}\)

Which is the correct solution.