Question

How do you integrate `int x sqrt( 3x^2 - 18x + 20 )dx` using trigonometric substitution?

Answer 1

Use the substitutions `sqrt3(x-3)=sqrt7u` and `u=sectheta`.

Explanation:

Let

`I=intxsqrt(3x^2-18x+20)dx`

Complete the square in the square root:

`I=intxsqrt(3(x-3)^2-7)dx`

Apply the substitution `sqrt3(x-3)=sqrt7u`:

`I=int(sqrt(7/3)u+3)(sqrt7sqrt(u^2-1))(sqrt(7/3)d theta)`

Integration is distributive:

`I=(7sqrt7)/3intusqrt(u^2-1)du+7sqrt3intsqrt(u^2-1)du`

The first integral is trivial. For the second apply the substitution `u=sectheta`:

`I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3intsecthetatan^2thetad theta`

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

`I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3int(sec^3theta-sectheta)d theta`

These are known integrals:

`I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3{1/2(secthetatantheta+ln|sectheta+tantheta|)-ln|sectheta+tantheta|}+C`

Simplify:

`I=(7sqrt7)/9(u^2-1)^(3/2)+(7sqrt3)/2(secthetatantheta-ln|sectheta+tantheta|)+C`

Reverse the last substitution:

`I=(7sqrt7)/9(u^2-1)^(3/2)+(7sqrt3)/2(usqrt(u^2-1)-ln|u+sqrt(u^2-1)|)+C`

Simplify:

`I=7/18(2sqrt7(u^2-1)+9sqrt3u)sqrt(u^2-1)-(7sqrt3)/2ln|u+sqrt(u^2-1)|+C`

Rewrite in terms of `sqrt7u` and rescale C:

`I=1/18(2(7u^2-7)+9sqrt3(sqrt7u))sqrt(7u^2-7)-(7sqrt3)/2ln|sqrt7u+sqrt(7u^2-7)|+C`

Reverse the first substitution:

`I=1/18(2(3x^2-18x+20)+27(x-3))sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C`

Simplify:

`I=1/18(6x^2-9x-41)sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C`

Answer 2

`I=1/9(3x^2-18x+20)^(3/2)+(3(x-3))/2sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C`

Explanation:

We know that,

#color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|x+sqrt(x^2- a^2)|+c#

`color(blue)((2)int[f(x)]^n f'(x)dx=[f(x)]^(n+1)/(n+1)+c`

NOTE : `color(violet)(ul{f(x)=3x^2-18x+20=>f'(x)=6x-18,`

#color(violet)(ul(so,take,x=1/6(6x)=1/6(6x-18+18)=[1/6(6x- 18)+3]#

Here,

`I=intxsqrt(3x^2-18x+20)dx`

`I=int[1/6(6x-18)+3]sqrt(3x^2-18x+20)dx`

=`int1/6(6x-18)sqrt(3x^2-18x+20)dx+3intsqrt(3x^2-18x+20)dx`

`I=I_1+I_2`

Now,

`I_1=int1/6(6x-18)sqrt(3x^2-18x+20)dx`

`=1/6color(blue)(int[3x^2-18x+20]^(1/2)d/(dx)(3x^2-18x+20)dx)`

`=1/6color(blue)((3x^2-18x+20)^(3/2)/(3/2)+c_1...toApply(2)`

`I_1=1/9(3x^2-18x+20)^(3/2)+c_1`

Again,

`I_2=3intsqrt(3x^2-18x+20)dx`

`=3intsqrt(3x^2-18x+27-7)dx`

`=3intsqrt(3(x^2-6x+9)-7)dx`

`I_2=3intsqrt(3(x-3)^2-(sqrt(7))^2)dx`

Subst. `sqrt3(x-3)=u=>sqrt3dx=du=>dx=1/sqrt3du`

`I_2=3/sqrt3color(red)(intsqrt(u^2-(sqrt7)^2)du...toApply(1)`

#=sqrt3color(red)({u/2sqrt(u^2-(sqrt7)^2)-(sqrt7)^2/2ln|u+sqrt(u^2- (sqrt7)^2)|})+c_2#

#=sqrt3{sqrt3/2(x-3)sqrt(3(x-3)^2-(sqrt7)^2)-7/2ln|sqrt3(x- 3)+sqrt(3(x-3)^2-(sqrt7)^2)|}+c_2#

#I_2=(3(x-3))/2sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x- 3)+sqrt(3x^2-18x+20)|+c_2#

Hence, `I=I_1+I_2=>`

#I=1/9(3x^2-18x+20)^(3/2)+(3(x-3))/2sqrt(3x^2- 18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C#