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

1 Answer
Jun 5, 2018

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

Explanation:

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