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

Jun 6, 2018

Use the substitutions $\sqrt{3} \left(x - 3\right) = \sqrt{7} u$ and $u = \sec \theta$.

#### Explanation:

Let

$I = \int x \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

Complete the square in the square root:

$I = \int x \sqrt{3 {\left(x - 3\right)}^{2} - 7} \mathrm{dx}$

Apply the substitution $\sqrt{3} \left(x - 3\right) = \sqrt{7} u$:

$I = \int \left(\sqrt{\frac{7}{3}} u + 3\right) \left(\sqrt{7} \sqrt{{u}^{2} - 1}\right) \left(\sqrt{\frac{7}{3}} d \theta\right)$

Integration is distributive:

$I = \frac{7 \sqrt{7}}{3} \int u \sqrt{{u}^{2} - 1} \mathrm{du} + 7 \sqrt{3} \int \sqrt{{u}^{2} - 1} \mathrm{du}$

The first integral is trivial. For the second apply the substitution $u = \sec \theta$:

$I = \frac{7 \sqrt{7}}{9} {\left({u}^{2} - 1\right)}^{\frac{3}{2}} + 7 \sqrt{3} \int \sec \theta {\tan}^{2} \theta d \theta$

Since ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$:

$I = \frac{7 \sqrt{7}}{9} {\left({u}^{2} - 1\right)}^{\frac{3}{2}} + 7 \sqrt{3} \int \left({\sec}^{3} \theta - \sec \theta\right) d \theta$

These are known integrals:

$I = \frac{7 \sqrt{7}}{9} {\left({u}^{2} - 1\right)}^{\frac{3}{2}} + 7 \sqrt{3} \left\{\frac{1}{2} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right) - \ln | \sec \theta + \tan \theta |\right\} + C$

Simplify:

$I = \frac{7 \sqrt{7}}{9} {\left({u}^{2} - 1\right)}^{\frac{3}{2}} + \frac{7 \sqrt{3}}{2} \left(\sec \theta \tan \theta - \ln | \sec \theta + \tan \theta |\right) + C$

Reverse the last substitution:

$I = \frac{7 \sqrt{7}}{9} {\left({u}^{2} - 1\right)}^{\frac{3}{2}} + \frac{7 \sqrt{3}}{2} \left(u \sqrt{{u}^{2} - 1} - \ln | u + \sqrt{{u}^{2} - 1} |\right) + C$

Simplify:

$I = \frac{7}{18} \left(2 \sqrt{7} \left({u}^{2} - 1\right) + 9 \sqrt{3} u\right) \sqrt{{u}^{2} - 1} - \frac{7 \sqrt{3}}{2} \ln | u + \sqrt{{u}^{2} - 1} | + C$

Rewrite in terms of $\sqrt{7} u$ and rescale C:

$I = \frac{1}{18} \left(2 \left(7 {u}^{2} - 7\right) + 9 \sqrt{3} \left(\sqrt{7} u\right)\right) \sqrt{7 {u}^{2} - 7} - \frac{7 \sqrt{3}}{2} \ln | \sqrt{7} u + \sqrt{7 {u}^{2} - 7} | + C$

Reverse the first substitution:

$I = \frac{1}{18} \left(2 \left(3 {x}^{2} - 18 x + 20\right) + 27 \left(x - 3\right)\right) \sqrt{3 {x}^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} \ln | \sqrt{3} \left(x - 3\right) + \sqrt{3 {x}^{2} - 18 x + 20} | + C$

Simplify:

$I = \frac{1}{18} \left(6 {x}^{2} - 9 x - 41\right) \sqrt{3 {x}^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} \ln | \sqrt{3} \left(x - 3\right) + \sqrt{3 {x}^{2} - 18 x + 20} | + C$

Jun 6, 2018

$I = \frac{1}{9} {\left(3 {x}^{2} - 18 x + 20\right)}^{\frac{3}{2}} + \frac{3 \left(x - 3\right)}{2} \sqrt{3 {x}^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} \ln | \sqrt{3} \left(x - 3\right) + \sqrt{3 {x}^{2} - 18 x + 20} | + C$

#### 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,

$I = \int x \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

$I = \int \left[\frac{1}{6} \left(6 x - 18\right) + 3\right] \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

=$\int \frac{1}{6} \left(6 x - 18\right) \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx} + 3 \int \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

$I = {I}_{1} + {I}_{2}$

Now,

${I}_{1} = \int \frac{1}{6} \left(6 x - 18\right) \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

$= \frac{1}{6} \textcolor{b l u e}{\int {\left[3 {x}^{2} - 18 x + 20\right]}^{\frac{1}{2}} \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 18 x + 20\right) \mathrm{dx}}$

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

${I}_{1} = \frac{1}{9} {\left(3 {x}^{2} - 18 x + 20\right)}^{\frac{3}{2}} + {c}_{1}$

Again,

${I}_{2} = 3 \int \sqrt{3 {x}^{2} - 18 x + 20} \mathrm{dx}$

$= 3 \int \sqrt{3 {x}^{2} - 18 x + 27 - 7} \mathrm{dx}$

$= 3 \int \sqrt{3 \left({x}^{2} - 6 x + 9\right) - 7} \mathrm{dx}$

${I}_{2} = 3 \int \sqrt{3 {\left(x - 3\right)}^{2} - {\left(\sqrt{7}\right)}^{2}} \mathrm{dx}$

Subst. $\sqrt{3} \left(x - 3\right) = u \implies \sqrt{3} \mathrm{dx} = \mathrm{du} \implies \mathrm{dx} = \frac{1}{\sqrt{3}} \mathrm{du}$

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, $I = {I}_{1} + {I}_{2} \implies$

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