How do you integrate #int x^3/sqrt(4x^2+8x+82) dx# using trigonometric substitution?

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Answer 1 · 2018-06-05T12:57:17

Use the substitution #x+1=sqrt78/2tantheta#.

Let

#I=intx^3/sqrt(4x^2+8x+82)dx#

Complete the square in the denominator:

#I=intx^3/sqrt(4(x+1)^2+78)dx#

Apply the substitution #x+1=sqrt78/2tantheta#:

#I=int(sqrt78/2tantheta-1)^3/(sqrt78sectheta)(sqrt78/2sec^2thetad theta)#

Let #a=sqrt78/2#:

#I=1/2int(a^3tan^3theta-3a^2tan^2theta+3atantheta-1)secthetad theta#

Since #tan^2theta=sec^2theta-1#:

#I=1/2int{a^3sec^3thetatantheta+(3a-a^3)secthetatantheta-3a^2sec^3theta+(3a^2-1)sectheta}d theta#

These are all known integrals:

#I=1/2{1/3a^3sec^3theta+(3a-a^3)sectheta-3/2a^2(secthetatantheta+ln|sectheta+tantheta|)+(3a^2-1)ln|sectheta+tantheta|}+C#

Rearrange and rescale C:

#I=1/12(2(atantheta)^2-9(atantheta)+18-4a^2)(asectheta)+1/8(6a^2-4)ln|atantheta+asectheta|+C#

Reverse the substitutions:

#I=1/12(2x^2-5x-67)sqrt((x+1)^2+39/2)+113/8ln|(x+1)+sqrt((x+1)^2+39/2)|+C#