Question

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

Answer 1

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`