U = COS dv = COS I dI du (n - 1) cos"-2 x dx U = sin x Sandy is off to a bad start, having made an error from which it is nearly impossible to recover. In one sentence, explain Sandy's error. Problem (2): Evaluate the following integrals. Note: some integrals may not require integration by parts. 2a cos(3x) dx / (c) .3 (a) 0. 2x tan-(x2) dx (e) dx (f) z(In 3z)2 dz (Hint: start with (d) / e3 tan(e) dx c* (b) / t² sin(3t) dt the u-sub u = In 3z.) %3D Problem (3): There are plenty of integrals whose evaluations require the applications of more than one technique of integration. Sometimes the steps are straightforward, sometimes not so straightforward. Each of the following integrals can be evaluated by first making a substitution and then applying a second technique of integration.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Parts D, E, and F

U = COS
dv
= COS I dI
du
(n - 1) cos"-2
x dx
U = sin x
Sandy is off to a bad start, having made an error from which it is nearly impossible to recover. In one
sentence, explain Sandy's error.
Problem (2): Evaluate the following integrals. Note: some integrals may not require integration by parts.
2a cos(3x) dx
/
(c)
.3
(a)
0.
2x tan-(x2) dx
(e)
dx
(f)
z(In 3z)2 dz (Hint: start with
(d) /
e3 tan(e) dx
c*
(b) / t² sin(3t) dt
the u-sub u = In 3z.)
%3D
Problem (3): There are plenty of integrals whose evaluations require the applications of more than one
technique of integration. Sometimes the steps are straightforward, sometimes not so straightforward. Each
of the following integrals can be evaluated by first making a substitution and then applying a second
technique of integration.
Transcribed Image Text:U = COS dv = COS I dI du (n - 1) cos"-2 x dx U = sin x Sandy is off to a bad start, having made an error from which it is nearly impossible to recover. In one sentence, explain Sandy's error. Problem (2): Evaluate the following integrals. Note: some integrals may not require integration by parts. 2a cos(3x) dx / (c) .3 (a) 0. 2x tan-(x2) dx (e) dx (f) z(In 3z)2 dz (Hint: start with (d) / e3 tan(e) dx c* (b) / t² sin(3t) dt the u-sub u = In 3z.) %3D Problem (3): There are plenty of integrals whose evaluations require the applications of more than one technique of integration. Sometimes the steps are straightforward, sometimes not so straightforward. Each of the following integrals can be evaluated by first making a substitution and then applying a second technique of integration.
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