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Calculus  II                                              Test #3                                   April 15, 2005

Name____________________         R.  Hammack                               Score ______
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(1) ∫  x/sec(x^2)dx = 1/2∫  cos(x^2)2x dx = 1/2sin(x^2) + C


(2) ∫  x sec^2(x)  dx = x tan(x)-∫ tan(x)  dx = x tan(x)-ln| sec(x) | + C


u = x            dv = sec^2(x)  du = dx      v = tan(x)



(3)  ∫ sec(x)  dx = ln| sec(x) + tan(x) | + C



(4)  ∫^ sin^(-1)(x)dx =
x sin^(-1)(x) - ∫^ x1/(1 - x^2)^(1/2) dx =
x sin^(-1)(x) + 1/2∫^ (1 - x^2)^(-1/2)(-2x ) dx =
x sin^(-1)(x) + (1 - x^2)^(1/2)+ C =
x sin^(-1)(x) + (1 - x^2)^(1/2)+ C

(Using interation by parts with the following substitutions:)
u = sin^(-1)(x)             &nbs ... ;               v = x



(5) ∫  1/(x^(1/2) 3^x^(1/2)) dx = -2∫ 3^(-x^(1/2)) -1/(2x^(1/2)) dx = -2∫ 3^u  du =

-2 /ln(3)3^u + C = -2 /ln(3)3^(-x^(1/2)) + C =-2 /(ln(3) 3^x^(1/2))+ C


u = -x^(1/2)  du = -1/(2x^(1/2))



(6)  ∫ sin^3(x) cos^3(x) dx = ∫sin^3(x) cos^2(x)cos(x) dx =

∫ sin^3(x) (1-sin^2(x))cos(x) dx = ∫u^3(1-u^2)du =∫u^3-u^5du =
u^4/4-u^6/6+C=sin^4(x)/4-sin^6(x)/6+ C



(7) ∫  e^x/(9 - e^(2x))^(1/2) dx = ∫  1/(3^2 - (e^x)^2)^(1/2)e^x dx =

∫  1/(3^2 - u^2)^(1/2)du =sin^(-1)(u/3) + C = sin^(-1)(e^x/3) + C


u = e^x  du = e^xdx



(8)   ∫_0^2  1/(x^2 + 4)  dx =   ∫_0^2  1/(x^2 + 2^2)  dx = [1/2tan^(-1)(x/2)] _0^2 = 1/2tan^(-1)(1) =π/8


(9) ∫ (x^3 + x^2 + 3x + 1)/(x^2   + 1)^2  dx = ∫ ( x   + 1)/(x^2   + 1)+(2 x  )/(x^2   + 1)^2  dx =

∫ ( x )/(x^2   + 1)dx + ∫1 /(x^2   + 1)dx + ∫(2 x  )/(x^2   + 1)^2  dx =
=1/2ln(x^2+2) + tan^(-1)(x) -1 /(x^2   + 1)+ C



(x^3 + x^2 + 3x + 1)/(x^2   + 1)^2 = (A x   + B)/(x^2   + 1) + ... (A x + B) (x^2 + 1) + C x + D  x^3 + x^2 + 3x + 1 = A x^3 + B x^2 + (A + C) x + (B + D)
A =1
B =1
C = 2
D = 0


(10)   ∫ 1/(x^2 - 1 )^(1/2)   dx =∫ 1/(sec^2(θ) - 1 )^(1/2) sec(θ)tan(θ) dx =

∫ (sec(θ) tan(θ))/tan^2(θ) ^(1/2)  dx = ∫sec(θ) dx = ln| sec(θ) + tan(θ) | + C
=ln| x+tan(sec^(-1)(x)) |+C=ln| x +(x^2 - 1 )^(1/2)| + C


x = sec(θ)  dx = sec (θ) tan (θ)



(11)   ∫_2^∞e^(-2x) dx =

Underscript[lim , l∞]∫_2^le^(-2x) dx =Underscript[lim , l∞][-e^(-2x) /2] _2^l=Underscript[lim , l∞][-1 /(2e^(2x))] _2^l=

Underscript[lim , l∞](-1 /(2e^(2l))+1 /(2e^4)) = 0 + 1 /(2e^4) = 1 /(2e^4)



(12)   ∫_0^13/x^(1/2) dx =
(Note: The integral is improper)


=Underscript[lim , l0^+]∫_l^13x^(-1/2) dx = Underscript[lim , l0^+][3x^(1/2)/(1/2)] _l^1 = Underscript[lim , l0^+][6x^(1/2)] _l^1 = Underscript[lim , l0^+](61^(1/2)-6l^(1/2)) = 6



(13) Underscript[lim , x⟶1](e^x - e)/(x^2 - 1) = Underscript[lim , x⟶1]e^x/(2x^2) = e/2



(14)   Underscript[lim , x⟶∞]x e^(-x)=Underscript[lim , x⟶∞]x /e^x=Underscript[lim , x⟶∞]1/e^x = 0