Integrate #x^3/sqrt((x^2+16))# dx using trig substitution?

2 Answers
Shwetank Mauria
Jan 28, 2018

#intx^3/sqrt(x^2+16)dx=1/3(sqrt(16+x^2)/x)^3-sqrt(16+x^2)/x+c#

Explanation:

Let #x=4tanu#, then #dx=4sec^2udu#

also #secu=sqrt(1+x^2/16)=sqrt(16+x^2)/x#

and #I=intx^3/sqrt(x^2+16)dx#

= #int(64tan^3x)/sqrt(16tan^2u+16)xx4sec^2udu#

= #int(64tan^3x)/(4secu)xx4sec^2udu#

= #int64tan^3usecudu#

= #64int(sec^2u-1)(secutanu)du#

Let #v=secu#. Then #dv=secutanudu#:

and #I=int(v^2-1)dv#

= #1/3v^3-v+C#

= #1/3sec^3u-secu+C#

= #1/3(sqrt(16+x^2)/x)^3-sqrt(16+x^2)/x+c#

Ananda Dasgupta
Jan 28, 2018

# = 4/3 sqrt{x^2 +4}(x^2-8)#

Explanation:

Try #x=4 tan theta#. Then #x = 4 sec^2theta d theta#
So
#int x^3/sqrt{x^2+16} dx = int {64 tan^3 theta times 4 sec^2theta}/{4sectheta}d theta #
#= 64 int tan^3theta sec theta d theta = 64 int tan^2theta tantheta sec theta d theta#
#=64 int (sec^2theta -1) d(sec theta) = 64 ({sec^3theta}/3 -sec theta) #
# = 64/3 sec theta (sec^2theta-3)=64/3 sqrt{1+x^2/16}(1+x^2/16-3)#
# = 1/3 sqrt{x^2 +16}(x^2-32)#

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