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) .

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 hI = 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 AI = 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 .

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 partsI = 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 .

 5) We use a simple substitutionI = 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 .

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 .

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 2I = 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 .

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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