Mathematics 134 – Intensive Calculus for Science 2 Discussion 2 – Practice on Integrals by Parts Answers February 14, 2006

Integrals

Group 1:

∫ x cos(4x) dx

Answer: Let u = x, dv = cos(4x) dx

\frac{x}{4} sin(4x) + \frac{1}{16} cos(4x) + C

Group 2:

∫ x(ln x)^{2} dx

Answer: Let u = (ln x)^{2}, dv = x dx to start. After two applications of parts you should get:

\frac{x^{2}}{2} (ln x)^{2} − \frac{x^{2}}{2} ln x + \frac{x^{2}}{4} + C

Group 3:

∫ x\sqrt{5x + 3} dy

Answer: Let u = x, dv = \sqrt{5x + 3} dx

\frac{2x}{15}(5x + 3)^{3/2} − \frac{4}{375}(5x + 3)^{5/2} + C

Group 4:

∫ θ^{2} sin(2θ) dθ

Answer: Let u = θ^{2}, dv = sin(2θ) dθ to start. After two applications of parts, you should get:

\frac{−θ^{2}}{2} cos(2θ) + \frac{θ}{2} sin(2θ) + \frac{1}{4} cos(2θ) + C

Group 5:

∫ t^{2}e^{−3t} dt

Answer: Let u = t^{2}, dv = e^{−3t} to start. After two applications of parts, you should get:

\frac{−t^{2}}{3} e^{−3t} − \frac{2t}{9}e^{−3t} − \frac{2}{27}e^{−3t} + C

Group 6:

∫ \frac{ln(x)}{x^{2}} dx

Answer: Let u = ln x, dv = x^{−2} dx

\frac{−1}{x} ln x − \frac{1}{x} + C

Group 7:

∫ x^{5} cos(x^{3}) dx

Hint: The cos(x^{3}) should go in dv, but what “else” do you need to be able to find v? Answer: Let u = x^{3} and dv = x^{2} cos(x^{3}) dx. The idea is to include the derivative of the x^{3} inside the cosine (up to a constant), so that you can find v = ∫ dv by substitution.

\frac{x^{3}}{3} sin(x^{3}) + \frac{1}{3} cos(x^{3}) + C

Group 8:

∫ arcsin(x)dx

Answer: Let u = arcsin(x) and dv = dx. The idea is to include the derivative of the x^{3} inside the cosine (up to a constant), so that you can find v = ∫ dv by substitution.

x arcsin(x) + \sqrt{1 − x^{2}} + C