How do you integrate int x^3/sqrt(4x^2+8x+82) dx using trigonometric substitution?
1 Answer
Use the substitution
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
I=int(sqrt78/2tantheta-1)^3/(sqrt78sectheta)(sqrt78/2sec^2thetad theta)
Let
I=1/2int(a^3tan^3theta-3a^2tan^2theta+3atantheta-1)secthetad theta
Since
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