MHB 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:

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

Let's try the substitution:

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

And we obtain:

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

Next, let's try integration by parts where:

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

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

And now we obtain:

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

For the remaining integral, consider:

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

Hence, we obtain:

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

From our original substitution, we find that:

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

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

And so, we finally find:

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