How do you integrate #int x^3 / ((sqrt(16+x^2))^3) dx# using trigonometric substitution?

Question

Luke Alvarez · Accepted Answer

#sqrt(16+x^2)+16/sqrt(16+x^2)#

Remember that #tan^2 theta+1=sec^2 theta#

For the case is convenient that

#x=4tany# #dx=4sec^2 y *dy#

Then

#int x^3/(sqrt(16+x^2))^3dx=int (64tan^3 y)/(sqrt(16+16tan^2 y))^3*4sec^2y* dy# #=int (256 tan^3 y*sec^2 y)/(4sec y)^3dy=int (4 tan^3 y)/sec y dy# #=4*int sin^3 y/cos^3 y*cos y*dy=4int sin y/cos^2 y*(1-cos^2 y) dy# #=4int sin y/cos^2 y*dy-4int sin y*dy# [A]

#-> int sin y/cos^2 y*dy=# Making #cos y=z# and #-sin y*dy=dz# #=- int dz/z^2=1/z=1/cos y#

Inserting the result above in expression [A]: #=4/cos y+4cos y +const.# [B]

But

#4tan y=x# => #siny=x/4*cos y# #-> cos^2 y+sin^2 y = 1# => #cos^2 y+x^2/16*cos^2 y=1# => #(1+x^2/16)cos^2 y=1# => #cos y=sqrt(16/(16+x^2))=4/sqrt(16+x^2)#

Inserting the result above in expression [B]:

#=cancel(4)/(cancel(4)/sqrt(16+x^2))+4*4/sqrt(16+x^2)+const.# #=sqrt(16+x^2)+16/sqrt(16+x^2)+const.#

Scarlett Allison · Answer

undefined To integrate $ \frac{x^3}{(\sqrt{16+x^2})^3} $ using trigonometric substitution, perform the following steps:

1. Substitute $ x = 4\tan(\theta) $.
2. Calculate $ dx $ using the derivative of $ \tan(\theta) $.
3. Substitute $ x $ and $ dx $ in terms of $ \theta $.
4. Simplify the integral in terms of $ \theta $.
5. Integrate with respect to $ \theta $.
6. Substitute back $ \theta $ in terms of $ x $ to get the final result.