University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2023
Views:

226

Pages:

3
Author:

Finn Hardin

Practice with Integrals Work out the following antiderivatives. There is a “hint sheet” attached. For the solutions, see the Maple file online. 1. 2. 3. 4. 5. Z Z Z Z Z 2x2 − x + 4 dx x3 + 4x 6. 2θ 7. e sin(3θ) dθ 1 dy y(2 − y) 8. t2 cos(3t) dt 9. x−1 dx x2 + 1 10. 1 Z y sinh(y) dy Z x4 dx − x2 Z t2 dt t+4 Z x−1 dx x+4 Z tan−1 (x) dx Hint Sheet: Integration Practice 1. 2x2 − x + 4 dx x3 + 4x Use Partial Fraction Decomposition, Z 2x2 − x + 4 A Bx + C = + 2 3 x + 4x x x +4 Once you’ve found A, B, C, break the integral up: A Z Z Z x 1 1 dx + B dx + C dx x x2 + 4 x2 + 4 Use u, du substitution for the second integral, and do some algebra on the last integral so you can use the inverse tangent as the antiderivative. 2. Z e2θ sin(3θ) dθ Integration by parts twice so that: Z Now solve for 3. 4 Z 2θ e sin(3θ) dθ = e (...) − e sin(3θ) dθ 9 2θ 2θ R 2θ e sin(3θ) dθ. 1 dy y(2 − y) Use partial fractions: Z 1 A B = + y(2 − y) y 2−y then integrate. You might write it as A/y − B/(y − 2) to make it easier to integrate later. 4. Z t2 cos(3t) dt Integration by parts; put t2 in the middle column of the table so that the third derivative is zero. 5. x−1 dx x2 + 1 We can write this as: Z Z x 1 dx − dx 2 2 x +1 x +1 Use u, du substitution for the first integral, the second integral is already in a nice form. Z 2 6. Z y sinh(y) dy Use integration by parts, where the antiderivative of sinh(y) is cosh(y), and the antiderivative of cosh(y) is sinh(y) (so put y in the middle column). 7. dx − x2 Factor the denominator, and use integration by parts. The x2 term can be thought of as a doubled up linear factor, so we would use: Z x4 A B + x x2 or as a single quadratic factor, in which case we would use: Ax + B x2 (These are equivalent). In any event, 1 x2 (x2 − 1) 8. 9. = A B C D + 2+ + x x x−1 x+1 t2 dt t+4 The degree of the numerator is greater than the degree of the denominator. Use long division to get: t2 16 = t2 − 4 + t+4 t+4 Z x−1 dx x+4 The degree of the numerator is larger than or equal to the degree of the denominator, so perform long division first: Z x−1 5 =1− x+4 x+4 10. Z tan−1 (x) dx Use integration by parts, with u = tan−1 (x) and dv = 1 dx. 3

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