MATH 132, section 1 – Calculus for Physical and Life Sciences Discussion 2 – Integration By Parts Practice February 4, 2008

The method of integration by parts, applies the “reverse product rule”

∫ u dv = uv − ∫ v du  (1)

to “trade in” a harder integral ∫ u dv for a hopefully simpler one ∫ v du. The steps for this method are

1. Split up the original integrand into the two parts u and dv. (Note: everything has to go into one part or the other – no “left-overs”)

2. From u compute du = u′(x)dx. From dv, compute v = ∫ dv.

3. Apply the reverse product rule (1)

4. Finish by computing ∫ v du. (This could require a substitution, or some other method.)

5. When there is a choice for the u and dv, the good choice will give ∫ v du that is simpler (or at least no more complicated than) ∫ u dv. If your ∫ v du is more complicated (cannot be simplified), then you might need to reconsider the choice of u, dv.

Do all the integrals below, and put the work for your group’s integral up on the board.

1. (I) and (V)

∫ x^{2} cos(3x) dx

2. (II) and (VI)

∫ x^{3}\sqrt{9 + x^{2}} dx

(Hint: The dv for this one should contain the \sqrt{9 + x^{2}}. What else do you need to be able to integrate dv by a method we have discussed?)

3. (III) and (VII)

∫ tan^{−1} 3x dx

4. (IV) and (VIII)

∫ x ln(x + 2) dx

(Hint: Substitute w = x + 2, then use parts.)