Calculus II Assignment # 3 |
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. |
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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 |
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. |
u_{1} = sin (ln x) du_{1} = cos (ln x) dx / x |
dv_{1} = dx v_{1} = 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 |
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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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