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Calculus  II                                                                 Test #3                                                              April 16, 2003

Name____________________                           R.  Hammack                                                            Score ______
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(1) ∫  ln(x)   dx =x ln(x) - x + C

(2) ∫  tan(x)   dx =-ln | cos(x) | +C

(3) ∫  sec(x^2) x  dx =1/2∫  sec(x^2) 2x  dx =1/2∫  sec(u)   du = 1/2ln | sec(u) + tan(u) | +C = 1/2ln | sec(x^2) + tan(x^2) | +C

u = x^2
du = 2x dx

(4) ∫  1/(9 + 4x^2) dx =1/2∫  1/(3^2 + (2x)^2) 2dx =1/2∫  1/(3^2 + u^2) du = 1/21/3tan^(-1)(u/3) + C = 1/6tan^(-1)((2x)/3) + C

u = 2x
du = 2 dx


(5) ∫  sin(3x)/(2 + cos(3x)) dx =-1/3∫   (-sin(3x) 3)/(2 + cos(3x)) dx =-1/3∫  1/udu = -1/3ln | u | +C = -1/3ln | 2 + cos(3x) | +C

u = 2 + cos(3x)
du = -sin(3x) 3 dx

(6)  ∫  x  e^(8x)   dx =(x e^(8x)   )/8 - ∫ 1/8e^(8x)   dx =(x e^(8x)   )/8 - 1/64e^(8x)    + C

u = x              dv = e^(8x)   dx
du = dx         v = 1/8e^(8x)   


(7)  ∫  cos^5(7x) sin(7x)   dx =-1/7∫  cos^5(7x) (-7) sin(7x)   dx =-1/7∫  u^5 du =-1/7u^6/6 + C =-cos^6(7x)/42 + C

u = cos(7x)
du = -sin(7x) 7dx


(8)  ∫  sec^5(x) tan^3(x)   dx =∫  sec^4(x) tan^2(x) sec(x) tan(x) dx =∫  sec^4(x) (sec^2(x) - 1) sec(x) tan(x) dx

Now make the substitution u = sec(x), du = sec(x) tan(x) dx  and this becomes

∫  u^4(u^2 - 1) du = ∫ u^6 - u^4 du = u^7/7 - u^5/5 + C = sec^7(x)/7 - sec^5(x)/5 + C


(9) ∫  x^(1/2) ln(x)   dx =ln(x) 2/3x^(3/2) - ∫  2/3x^(3/2) 1/x  dx =ln(x) 2/3x^(3/2) - 2/3∫  x^(1/2)   dx =(2ln(x) x^(1/2)^3)/3 - 4/9 x^(1/2)^3 + C


u = ln(x)               dv = x^(1/2)   dx
du = 1/x dx         v = 2/3x^(3/2)



(10) ∫ (16 - x^2)^(1/2) dx =

Use the following substitution:
x = 4sin(θ)
dx = 4cos(θ) dθ

∫ (16 - x^2)^(1/2) dx =∫ (16 - (4sin(θ))^2)^(1/2) 4cos(θ) dθ =∫ (16 - (4sin(θ))^2)^(1/2) 4cos(θ) dθ =∫ 4cos(θ) 4cos(θ) dθ =16∫cos^2(θ) dθ =16 (1/2θ + 1/2cos(θ) sin(θ)) = 8θ + 8cos(θ) sin(θ) + C

Now use the fact that you have sin(θ) = x/4 so θ = sin^(-1)(x/4)  and cos(θ) = (16 - x^2)^(1/2)/4 and the final answer is
8sin^(-1)(x/4) + 8 (16 - x^2)^(1/2)/4x/4 + C = 8sin^(-1)(x/4) + (x (16 - x^2)^(1/2))/2 + C

(11) ∫   (3x - 5)/(x^2 - 2x - 3) =

(3x - 5)/(x^2 - 2x - 3) = (3x - 5)/((x + 1) (x - 3)) = A/(x + 1) + B/(x - 3)

3x - 5 = A(x - 3) + B(x + 1)

To find A, set x = -1:
3 (-1) - 5 = A(-1 - 3) + B(-1 + 1)
-8 = -4A
A = 2

To find B, set x = 3:
3 (3) - 5 = A(3 - 3) + B(3 + 1)
4 = 4B
B = 1

∫   (3x - 5)/(x^2 - 2x - 3) =∫  2/(x + 1) dx + ∫  1/(x - 3) dx = 2ln | x + 1 | +ln | x - 3 | +C = ln | (x + 1)^2 (x - 3) | +C



(12) Underscript[lim , x⟶∞] ln(x)/x =Underscript[lim , x⟶∞] (1/x)/1 = 0



(13)   Underscript[lim , x⟶0] (1 + 2x)^(-3/x) =Underscript[lim , x⟶0] e^ln(1 + 2x)^(-3/x) =Underscript[lim , x⟶0] e^(-3/xln(1 + 2x)) =Underscript[lim , x⟶0] e^(-3ln(1 + 2x))/x

Now the power of e is of indeterminate form 0/0 so we use L'Hopital's Rule.

Underscript[lim , x⟶0] e^(-32/(1 + 2x))/1 = e^(-6) = 1/e^6