Integral of (x+3)sqrt[(3-4x-x^2)]dx?

Hello,

∫ (x + 3) √(3 - 4x - x²) dx =

let's complete the square inside the root by adding and subtracting 4:

∫ (x + 3) √(3 + 4 - 4 - 4x - x²) dx =

∫ (x + 3) √[7 - (x² + 4x + 4)] dx =

∫ (x + 3) √[7 - (x + 2)²] dx =

(rewriting the radicand as a difference of two squares)

∫ (x + 3) √[(√7)² - (x + 2)²] dx =

let:

x + 2 = (√7)sinθ

x = (√7)sinθ - 2

dx = (√7)cosθ dθ

yielding, by trig substitution:

∫ (x + 3) √[(√7)² - (x + 2)²] dx = ∫ {[(√7)sinθ - 2] + 3} √{(√7)² - [(√7)sinθ]²} dx =

∫ [(√7)sinθ - 2 + 3] [√(7 - 7sin²θ)] (√7)cosθ dθ =

∫ [(√7)sinθ + 1] {√[7(1 - sin²θ)]} (√7)cosθ dθ =

(replacing 1 - sin²θ with cos²θ)

∫ [(√7)sinθ + 1] [√(7cos²θ)] (√7)cosθ dθ =

∫ [(√7)sinθ + 1] (√7)cosθ (√7)cosθ dθ =

∫ [(√7)sinθ + 1] (√7)²cos²θ dθ =

∫ [(√7)sinθ + 1] 7cos²θ dθ =

∫ [7(√7)cos²θ sinθ + 7cos²θ] dθ =

let's rewrite the argument of the second cos²θ as (2θ)/2:

∫ {7(√7)cos²θ sinθ + 7cos²[(2θ)/2]} dθ =

let's apply (to cos²[(2θ)/2]) the half-angle formula cos²(α/2) = (1 + cos α)/2:

∫ {7(√7)cos²θ sinθ + 7{[1 + cos(2θ)] /2} } dθ =

∫ {7(√7)cos²θ sinθ + (7/2)[1 + cos(2θ)]} dθ =

∫ [7(√7)cos²θ sinθ + (7/2) + (7/2)cos(2θ)] dθ =

(splitting into three integrals and factoring constants out)

7(√7) ∫ cos²θ sinθ dθ + (7/2) ∫ dθ + (7/2) ∫ cos(2θ) dθ =

(dividing and multiplying the last integral by 2 that is the derivative of the argument 2θ)

7(√7) ∫ cos²θ sinθ dθ + (7/2) ∫ dθ + (7/2)(1/2) ∫ cos(2θ) 2 dθ =

7(√7) ∫ cos²θ sinθ dθ + (7/2)θ + (7/4) ∫ cos(2θ) d(2θ) =

(applying the integration rule ∫ cos[f(x)] d[f(x)] = sin[f(x)] + C)

7(√7) ∫ cos²θ sinθ dθ + (7/2)θ + (7/4)sin(2θ) =

let's change signs in the remaining integrand:

7(√7) ∫ (- cos²θ) (- sinθ) dθ + (7/2)θ + (7/4)sin(2θ) =

(being - sinθ the derivative of cosθ)

7(√7) ∫ (- cos²θ) d(cosθ) + (7/2)θ + (7/4)sin(2θ) =

- 7(√7) ∫ cos²θ d(cosθ) + (7/2)θ + (7/4)sin(2θ) =

(by the integration rule ∫ f(x)ⁿ d[f(x)] = [1/(n+1)] f(x)ⁿ⁺¹ + C)

- 7(√7) [1/(2+1)] (cosθ)² ⁺ ¹ + (7/2)θ + (7/4)sin(2θ) + C =

- 7(√7) (1/3)(cosθ)³ + (7/2)θ + (7/4)sin(2θ) + C =

(applying the double-angle formula sin(2θ) = 2sinθ cosθ)

- [(7√7)/3]cos³θ + (7/2)θ + (7/4)2sinθ cosθ + C =

- [(7√7)/3]cos³θ + (7/2)θ + (7/2)sinθ cosθ + C

let's now recall that x + 2 = (√7)sinθ; hence:

sinθ = (x + 2)/√7

θ = arcsin[(x + 2)/√7]

cosθ = √(1 - sin²θ) = √{1 - [(x + 2)/√7]²} = √{1 - [(x + 2)²/7]} =

√{1 - [(x² + 4x + 4)/7]} =

√{[7 - (x² + 4x + 4)} /7} =

√[(7 - x² - 4x - 4)} /7] =

[√(3 - 4x - x²)] /√7

then, substituting back:

- [(7√7)/3]cos³θ + (7/2)θ + (7/2)sinθ cosθ + C = - [(7√7)/3] {[√(3 - 4x - x²)] /√7}³ +

(7/2)arcsin[(x + 2)/√7] + (7/2) [(x + 2)/√7] {[√(3 - 4x - x²)] /√7} + C =

- [(7√7)/3] {[√(3 - 4x - x²)³] /√7³} + (7/2)arcsin[(x + 2)/√7] + (7/2) [(x +

2)/√7²] √(3 - 4x - x²) + C =

- [(7√7)/3] {[√(3 - 4x - x²)³] /√(7² ∙ 7)} + (7/2)arcsin[(x + 2)/√7] + (7/2)[(x +

2)/7] √(3 - 4x - x²) + C =

- [(7√7)/3] {[√(3 - 4x - x²)³] /(7√7)} + (7/2)arcsin[(x + 2)/√7] + (1/2)(x + 2)√(3 -

4x - x²) + C =

ending with:

- (1/3)√(3 - 4x - x²)³ + (7/2)arcsin[(x + 2)/√7] + (1/2)(x + 2)√(3 - 4x - x²) + C =

I hope it's helpful

f(x)=(x+3)sqrt[7-(x+2)^2]

=(x+2+1)sqrt[7-(x+2)]

=(x+2)sqrt[7-(x+2)^2] + sqrt[7-(x+2)^2]

integralf(x)dx = M+N where

+M=integral(x+2)*[7-(x+2)^2]^(1/2)dx

by substitute: u=(x+2)^2 -->du=2(x+2)dx

-->M=(1/2)integral[(1-u)^(/1/2)]du

=(1/4)*(1-u)^(-1/2)

=1/[4sqrt(1-u)]

=1/[4sqrt(1-(x+2)^2] +c

+ N=integral[sqrt(7- (x+2)^2]dx

by substitute

x+2=(sqrt7)sint

--->dx=(sqrt7)cost.dt

and t=arcsin[(x+2)/sqrt7]

-->N=(sqrt7)integral[(cost)^2.dt

=(sqrt7/2)integral[1+cos2t]dt

=(sqrt7/2)[t +(1/2)sin2t]+ c

=(sqrt7/2)[ t +sintcost] +c

= ------------

at last: integral=M+N