How do you integrate #int 1/(x^4sqrt(16+x^2))# by trigonometric substitution?

Question

Charles Arocha · Accepted Answer

The answer is #=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C#

Let #x=4tantheta#, #=>#, #dx=4sec^2thetad theta#

#sqrt(16+x^2)=sqrt(16+16tan^2theta)=4sectheta#

Therefore, the integral is

#I=int(dx)/(x^4sqrt(16+x^2))=int(4sec^2thetad theta)/(256tan^4theta*4sectheta)#

#=1/256int(secthetad theta)/(tan^4theta)#

#tantheta=sintheta/costheta#

#sectheta=1/costheta#

#I=1/256int(d theta)/(costheta*sin^4theta/cos^4 theta)#

#=1/256int(cos^3thetad theta)/(sin^4theta)#

#=1/256int(costheta(1-sin^2theta)d theta)/sin^4theta#

Let #v=sintheta#, #=>#, #dv=costheta d theta#

#I=1/256int((1-v^2)dv)/(v^4)#

#=1/256int(1/v^4-1/v^2)dv#

#=-1/256*1/(3v^3)+1/256*(1/v)#

#=1/256*1/sintheta-1/768*1/sin^3theta#

#=1/256*1/(x/sqrt(16+x^2))-1/768*1/(x^3/(x^2+16)^(3/2))+C#

#=sqrt(16+x^2)/(256x)-(x^2+16)^(3/2)/(768x^3)+C#

Evelyn Agard · Answer

undefined To integrate $ \int \frac{1}{x^4 \sqrt{16 + x^2}} $ by trigonometric substitution:

1. Substitute $ x = 4\tan(\theta) $.
2. Calculate $ dx = 4\sec^2(\theta) \, d\theta $.
3. Express $ \sqrt{16 + x^2} $ in terms of $ \theta $.
4. Replace $ x^4 $ with $ (4\tan(\theta))^4 $.
5. Substitute $ dx $ and the expressions for $ x $ and $ \sqrt{16 + x^2} $ into the integral.
6. Simplify the integral in terms of $ \theta $.
7. Integrate with respect to $ \theta $.
8. Convert back to the original variable $ x $ if needed.
9. Simplify the result.