SOLUTIONS FOR CALCULUS II TEST #2

A/ Integrate:

 1) I = ò x² ln x dx u = ln x dv = x² ò u dv = uv - ò v du du = dx / x v = 1/3 x³ I = 1/3 x³ ln x - 1/3 ò x² dx I = 1/3 x³ ln x - 1/9 x³ + C

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 2) I = ò e x cox x dx u = e x dv = cos x dx du = e x dx v = sin x I = e x sin x - ò e x sin x dx u1 = e x dv1 = sin x dx I = e x sin x - [ e x cox x + ò e x cox x dx] du1 = e x dx v1 = -cos x 2I = e x sin x - e x cox x I = ½ (e x sin x - e x cox x) + C

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 3) I = Let x = tan h, so dx = sec 2 h dh,

Now we look at our image of h

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4)

We let x = 3 sin h (since a 2 = 9)
This makes dx = 3 cos h dh, and it makes 9 - x 2 = 9 - 9 sin 2 h
Also, x 2 = 9 sin 2
h

Now the integral becomes

Now we look at our image of h

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 5) Let u = x² - 1, therefore, du = 2x dxso, ½du = x dxwhen x = 1, u = 0 and when x = 2, u = 3

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6)

We let x = 5 sec h
dx = 5 sec h tan h dh, x 2= 25 sec 2 h
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ò (sech - cos h ) dh = ln | sech + tan h | - sin h.

Now we look at our image of h

Returning to the original variable x we get

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7) ° sin ² x cos ³ x dx , odd power of cosine.
° sin ² x cos x cos ² x dx =°( sin ² x - sin 4 x) cos x dx = 1/3 sin³ x - 1/5 sin 5 x + C

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8) ° tan ³ x sec x dx = ° tan² x sec x tan x dx = ° (sec² x - 1) sec x tan x dx

= 1/3 sec³ x - sec x + C

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9)

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10) A = ½ = C, B = -½

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B/ 1)

A = 4y² = 4 (9 - x²)

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2)

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