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 dx so, ½du = x dx when 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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