Unformatted text preview: MATH 2202 — Exam 1 (Version 2)
January 29, 2007 S. F. Ellermeyer Name Instructions. This exam contains ﬁve indeﬁnite integrals (with answers given). On each
problem you must show a) how to evaluate the integral (using the strategies we have been
studying such as basic substitution, integration by parts, trigonometric substitution, etc.) and b)
how to check that the answer you get is correct (by differentiation). An example of how your work should appear is given here:
Example: Ixsec2(x2)dx = L 2
2 tan(x )+ C. a) To evaluate the above integral, we use a basic substitution: Let u = x2. Then du = Zxdx and we have J-xsecz(x2)dx = % Isecz(u)du = Ltan(u) + C = i tan(x2) + C. 2 2 b) To check that this result is correct, we observe that ii 22l22__d_2=L22,=.2.2
dx(2tan(x )) 2sec (x) dx(x) 2sec (x) 2x asec (x ). Some of the following trigonometric identities will be needed for certain problems: sin2(6) + cosz(0) : l sin(9) cos(6) = % sin(26) tan2(6) + 1 : sec2(9) cosz(0) = i0 + cos(20)) 1 + 009(0) = csc2(0) sin2(0) : é—(l — cos(29)) 1.
dex = J—(arcsin(x))2 + C.
’1 _ x2 2
[ashram = %ex(sin(x) — cos(x)) + C.
Isinzbc) cos2(x)dx = %x — 31—2 sin(4x) + C (There are other possible ways to express the answer on this one — depending on which trigonometric identities are used in doing the integral. Thus it is possible that the answer you get might look different from the above answer. If so, that is ﬁne, as long as you show in detail the method that you use. Also, you do not need to show the check on this one. Just compute the integral.)
4. L z ' (L _ i / _ 2
JIM dx 2arcsm 2x) 2x 4 x +C. 5. ...
View Full Document
- Fall '08