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Calculus II                                                Quiz #5                            March 26, 2004

Name____________________          R.  Hammack                         Score ______
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(1)      ∫e^x/(4 - e^(2x))^(1/2) dx =
∫e^x/(2^2 - (e^x)^2)^(1/2) dx = ∫1/(2^2 - u^2)^(1/2) du = sin^(-1)(u/2) + C = sin^(-1) (e^x/2) + C



(2)      ∫tan(5x) dx =

1/5∫tan(5x) 5dx = 1/5∫tan(u) 5du = 1/5ln | sec(u) | +C = 1/5ln | sec(5x) | +C



(3)      ∫e^xsin(x) dx =-cos(x) e^x - ∫ (-cos(x)) e^x dx = -cos(x) e^x + ∫e^xcos(x) dx

u = e^x              &nb ... ;              v = -cos (x)

Now we use integration by parts a second time.

u = e^x              &nb ... p;              v = sin (x)

∫e^xsin(x) dx = -cos(x) e^x + ∫e^xcos(x) dx = -cos(x) e^x + e^xsin(x) - ∫sin(x) e^xdx

Now, the above expression says
∫e^xsin(x) dx = -cos(x) e^x + e^xsin(x) - ∫e^xsin(x) dx.
From this we get
2∫e^xsin(x) dx = -cos(x) e^x + e^xsin(x)

so ∫e^xsin(x) dx = e^x(sin(x) - cos(x))/2 + C


(4)      ∫x^(1/2) ln(x) dx = ln(x) 2/3x^(3/2) - ∫2/3x^(3/2) 1/x dx = (2ln(x) x^(1/2)^3)/3 - 2/3∫x^(1/2) dx = (2ln(x) x^(1/2)^3)/3 - 2/32/3x^(3/2) + C =

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


u = ln(x)                ... nbsp;             v = 2/3x^(3/2)

Created by Mathematica  (March 26, 2004)