This assignment covers

Techniques of Integration including:

Integration by Parts

Trig Integrals

Trig Substitutions

Partial Fraction Decomposition

Miscellaneous Substitutions

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Questions

Use any or all techniques of Integration to evaluate

1)	2)
3)	4)
5)	6)
7)	8)
9)	10)
11)	12)

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Solutions

1)

I =, using integration by parts	u = arcsin x	dv = x dx
I = ½ x ² arcsin x -		v = ½ x ²
Now a trig sub on the integral	x = sin h	dx = cos h dh
I = ½ x ² arcsin x -
Now use the half-angle formula on sin² h I = ½ x ² arcsin x - I = ½ x ² arcsin x - ¼ (h - ½ sin 2h ) Now use the identity sin 2A = 2 sin A cos A I = ½ x ² arcsin x - ¼ (h - sin h cos h ) I = ½ x ² arcsin x -	from the trig sub we get the triangle

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2)

I = using integration by parts and ignoring the limits of integration until end.
I = x ln (1 + x) - Now divide x by x + 1	u = ln (1 + x) du = dx / (1 + x)	dv = dx v = x
I = x ln (1 + x) -
I = x ln | x + 1| - x + ln | x + 1| = 2 ln 2 - 1

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3)

I =first a substitution, then parts I =	let so y² = x, and 2y dy = dx
Now we use parts on the new integral
I =	u = y du = dy	dv = e y dy v = e y

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4) We use the "odd power of cosine" approach.

Now we set u = sin 2x and du / 2 = cos 2x dx, so the integral becomes

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5) We use a simple substitution I = Since u = 1 + cos x, I =

Let u = 1 + cos x, so - du = sin x dx

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6)

Now we apply the half-angle formula again to cos ² 2x, so the integral becomes

Integrating this expression gives us

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7) Treat this as odd power of tan x.

when we set u = sec x , and du = sec x tan x dx. .

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8)

We use a trig sub on the integral	x = 4 sin h	dx = 4 cos h dh
I = This integrates to 1/16 cot h which we get from the triangle created by the trig sub.	from the trig sub we get the triangle

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9)

I = using integration by parts
I = x cos (ln x) + Now we use parts again	u = cos (ln x) du = - sin (ln x) dx / x	dv = dx v = x
I = x cos (ln x) + x sin (ln x) which is I -- so this is a boomerang integral.	u1 = sin (ln x) du1 = cos (ln x) dx / x	dv1 = dx v1 = x
2I = x cos (ln x) + x sin (ln x), divide by 2 I = ½ ( x cos (ln x) + x sin (ln x)) + C

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10)

Evaluate

When we complete the square on the quadratic denominator we get
x² + 6x + 9 - 9 + 13 = (x + 3)² + 4, so now the integral is

so we use a substitution

Let u = x + 3, so du = dx and x = u - 3

Now our integral is

Using the substitution w = u² + 4, for the 1st integral, we get

½ ln (u² + 4) - (3/2) arcTan (u/2) and now we "unsubstitute"

The final answer is

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11) We factor the denominator and apply partial fraction decompostion.

I = ln | x | - ½ ln | x² + 1 | + K

The numerator on the right side is A(x² + 1) + (Bx + C) x = 1 Set x = 0 to get A = 1 then balance coefficients. We get B = - 1 and C = 0.

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12)

We use a miscellaneous substitution Now we use long division on the fraction When we integrate and unsubstitute we get

Let u = x 1 / 3 so u³ = x and 3 u² du = dx.

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