Integrate sqrt(1-x^4)/x^5 dx Using Trig Sub | Yahoo Answers

[MarkFL]

Here is the question:

How do I integrate sqrt(1-x^4)/x^5 dx using trig sub?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello The_Reporter,

We are given to compute the indefinite integral:

\(\displaystyle I=\int\frac{\sqrt{1-x^4}}{x^5}\,dx\)

Let's try the substitution:

\(\displaystyle x^2=\cos(\theta)\,\therefore\,dx=-\frac{\sin(\theta)}{2x}\,d\theta\)

And we obtain:

\(\displaystyle I=-\frac{1}{2}\int\frac{\sin^2(\theta)}{\cos^3(\theta)}\,d\theta\)

Next, let's try integration by parts where:

\(\displaystyle u=\sin(\theta)\,du=\cos(\theta)\,d\theta\)

\(\displaystyle dv=\frac{\sin(\theta)}{\cos(\theta}\,d\theta\,\therefore\,v=\frac{1}{2\cos^2(\theta)}\)

And now we obtain:

\(\displaystyle I=-\frac{1}{4}\left(\frac{\sin(\theta)}{\cos^2(\theta)}-\int\sec(\theta)\,d\theta\right)\)

For the remaining integral, consider:

\(\displaystyle \sec(\theta)=\frac{\sec(\theta)\left(\sec(\theta)+\tan(\theta)\right)}{\sec(\theta)+\tan(\theta)}=\frac{d}{d\theta}\left(\ln\left|\sec(\theta)+\tan(\theta)\right|\right)\)

Hence, we obtain:

\(\displaystyle I=-\frac{1}{4}\left(\frac{\sin(\theta)}{\cos^2(\theta)}-\ln\left|\sec(\theta)+\tan(\theta)\right|\right)+C\)

From our original substitution, we find that:

\(\displaystyle \tan(\theta)=\frac{\sqrt{1-x^4}}{x^2}\)

\(\displaystyle \sec(\theta)=\frac{1}{x^2}\)

And so, we finally find:

\(\displaystyle \bbox[5px,border:2px solid #207498]{I=\frac{1}{4}\left(\ln\left(\frac{\sqrt{1-x^4}+1}{x^2} \right)-\frac{\sqrt{1-x^4}}{x^4} \right)+C}\)