How do you integrate #int x^3 / ((sqrt(16+x^2))^3) dx# using trigonometric substitution?
1 Answer
Explanation:
Remember that
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.#