How do you integrate $\int x \sqrt{3 x^{2} - 18 x + 20} d x$ $int x sqrt( 3x^2 - 18x + 20 )dx$ using trigonometric substitution?

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How do you integrate $\int x \sqrt{3 x^{2} - 18 x + 20} d x$ $int x sqrt( 3x^2 - 18x + 20 )dx$ using trigonometric substitution?

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Answer 1 · 2018-06-06T10:06:41

Use the substitutions $\sqrt{3} (x - 3) = \sqrt{7} u$ $sqrt3(x-3)=sqrt7u$ and $u = sec θ$ $u=sectheta$ .

Explanation:

Let

$I = \int x \sqrt{3 x^{2} - 18 x + 20} d x$ $I=intxsqrt(3x^2-18x+20)dx$

Complete the square in the square root:

$I = \int x \sqrt{3 {(x - 3)}^{2} - 7} d x$ $I=intxsqrt(3(x-3)^2-7)dx$

Apply the substitution $\sqrt{3} (x - 3) = \sqrt{7} u$ $sqrt3(x-3)=sqrt7u$ :

$I = \int (\sqrt{\frac{7}{3}} u + 3) (\sqrt{7} \sqrt{u^{2} - 1}) (\sqrt{\frac{7}{3}} d θ)$ $I=int(sqrt(7/3)u+3)(sqrt7sqrt(u^2-1))(sqrt(7/3)d theta)$

Integration is distributive:

$I = \frac{7 \sqrt{7}}{3} \int u \sqrt{u^{2} - 1} d u + 7 \sqrt{3} \int \sqrt{u^{2} - 1} d u$ $I=(7sqrt7)/3intusqrt(u^2-1)du+7sqrt3intsqrt(u^2-1)du$

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

$I = \frac{7 \sqrt{7}}{9} {(u^{2} - 1)}^{\frac{3}{2}} + 7 \sqrt{3} \int sec θ {tan}^{2} θ d θ$ $I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3intsecthetatan^2thetad theta$

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

$I = \frac{7 \sqrt{7}}{9} {(u^{2} - 1)}^{\frac{3}{2}} + 7 \sqrt{3} \int ({sec}^{3} θ - sec θ) d θ$ $I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3int(sec^3theta-sectheta)d theta$

These are known integrals:

$I = \frac{7 \sqrt{7}}{9} {(u^{2} - 1)}^{\frac{3}{2}} + 7 \sqrt{3} {\frac{1}{2} (sec θ tan θ + ln | sec θ + tan θ |) - ln | sec θ + tan θ |} + C$ $I=(7sqrt7)/9(u^2-1)^(3/2)+7sqrt3{1/2(secthetatantheta+ln|sectheta+tantheta|)-ln|sectheta+tantheta|}+C$

Simplify:

$I = \frac{7 \sqrt{7}}{9} {(u^{2} - 1)}^{\frac{3}{2}} + \frac{7 \sqrt{3}}{2} (sec θ tan θ - ln | sec θ + tan θ |) + C$ $I=(7sqrt7)/9(u^2-1)^(3/2)+(7sqrt3)/2(secthetatantheta-ln|sectheta+tantheta|)+C$

Reverse the last substitution:

$I = \frac{7 \sqrt{7}}{9} {(u^{2} - 1)}^{\frac{3}{2}} + \frac{7 \sqrt{3}}{2} (u \sqrt{u^{2} - 1} - ln ∣ ∣ u + \sqrt{u^{2} - 1} ∣ ∣) + C$ $I=(7sqrt7)/9(u^2-1)^(3/2)+(7sqrt3)/2(usqrt(u^2-1)-ln|u+sqrt(u^2-1)|)+C$

Simplify:

$I = \frac{7}{18} (2 \sqrt{7} (u^{2} - 1) + 9 \sqrt{3} u) \sqrt{u^{2} - 1} - \frac{7 \sqrt{3}}{2} ln ∣ ∣ u + \sqrt{u^{2} - 1} ∣ ∣ + C$ $I=7/18(2sqrt7(u^2-1)+9sqrt3u)sqrt(u^2-1)-(7sqrt3)/2ln|u+sqrt(u^2-1)|+C$

Rewrite in terms of $\sqrt{7} u$ $sqrt7u$ and rescale C:

$I = \frac{1}{18} (2 (7 u^{2} - 7) + 9 \sqrt{3} (\sqrt{7} u)) \sqrt{7 u^{2} - 7} - \frac{7 \sqrt{3}}{2} ln ∣ ∣ \sqrt{7} u + \sqrt{7 u^{2} - 7} ∣ ∣ + C$ $I=1/18(2(7u^2-7)+9sqrt3(sqrt7u))sqrt(7u^2-7)-(7sqrt3)/2ln|sqrt7u+sqrt(7u^2-7)|+C$

Reverse the first substitution:

$I = \frac{1}{18} (2 (3 x^{2} - 18 x + 20) + 27 (x - 3)) \sqrt{3 x^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} ln ∣ ∣ \sqrt{3} (x - 3) + \sqrt{3 x^{2} - 18 x + 20} ∣ ∣ + C$ $I=1/18(2(3x^2-18x+20)+27(x-3))sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C$

Simplify:

$I = \frac{1}{18} (6 x^{2} - 9 x - 41) \sqrt{3 x^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} ln ∣ ∣ \sqrt{3} (x - 3) + \sqrt{3 x^{2} - 18 x + 20} ∣ ∣ + C$ $I=1/18(6x^2-9x-41)sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C$

Answer 2 · 2018-06-06T11:31:08

$I = \frac{1}{9} {(3 x^{2} - 18 x + 20)}^{\frac{3}{2}} + \frac{3 (x - 3)}{2} \sqrt{3 x^{2} - 18 x + 20} - \frac{7 \sqrt{3}}{2} ln ∣ ∣ \sqrt{3} (x - 3) + \sqrt{3 x^{2} - 18 x + 20} ∣ ∣ + C$ $I=1/9(3x^2-18x+20)^(3/2)+(3(x-3))/2sqrt(3x^2-18x+20)-(7sqrt3)/2ln|sqrt3(x-3)+sqrt(3x^2-18x+20)|+C$

Explanation:

We know that,

$(1) \int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} ln ∣ ∣ x + \sqrt{x^{2} - a^{2}} ∣ ∣ + c$ $color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|x+sqrt(x^2- a^2)|+c$

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

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, $I=I_1+I_2=>$

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

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2 Answers

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