MATH 1172 Midterm: MATH 1172 Ohio State University Math 1172 7.2 Solutions Fa 2014
31 Jan 2019
School
Department
Course
Professor
Math 1172 (Buenger) Section 7.2 September 15, 2015
1.
Z2xe3xdx
Solution: Let
u= 2xand dv =e3xdx.
Then
du = 2 dx and v=1
3e3x.
Z2x e3xdx = 2x·1
3e3x−Z21
3e3xdx
= 2x·1
3e3x−2
3·1
3e3x+C.
2.
Zxln x dx
Solution: Let
u= ln(x) and dv =x dx.
Then
du =1
xdx and v=x2
2.
So
Zxln x dx =x2
2ln(x)−Z1
x·x2
2dx
=x2
2ln(x)−Zx
2dx
=x2
2ln(x)−x2
4+C.
3.
Zxsin xcos x dx
Solution: Let
u=xand dv = sin xcos x dx =1
2sin(2x).a
Thus
du =dx and v=−1
4cos(2x).
So
Zxsin xcos x
=x·−1
4cos(2x)−Z−1
4cos(2x)dx
=−x
4cos(2x) + 1
8sin(2x)dx.
aRecall that sin xcos x=1
2sin(2x).
4.
Ztan−1xdx
Solution: Let
u= tan−1(x) and dv =dx.
Then
du =1
1 + x2dx and v=x.
So
Ztan−1xdx =xtan−1(x)−Zx
1 + x2dx
=xtan−1(x)−1
2Z1
wdw
=xtan−1(x)−1
2ln |1 + x2|dw
5.
Z1
0
sin (2x)exdx
1
4
MATH 1172 Full Course Notes
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Document Summary
Solution: let u = 2x and dv = e3x dx. Solution: let u = ln(x) and dv = x dx. 2 ln(x) z 1 ln(x) z x x. Solution: let u = x and dv = sin x cos x dx = 8 sin(2x) dx. arecall that sin x cos x = 1. Solution: let u = tan 1(x) and dv = dx. Solution: let us rst nd the inde nite inte: (test question) gral r sin(2x)ex dx. Let u = sin(2x) and dv = ex dx. Then du = 2 cos(2x) dx and v = ex. Z sin(2x)ex dx = sin(2x)ex z 2 cos(2x)ex dx. Solution: let u = x2 and dv = e3x dx. U = cos(2x) and d v = ex dx. Z sin(2x)ex dx = sin(2x)ex 2hex cos(2x) = sin(2x)ex 2ex cos(2x) above equation, we nd that. By adding 4r sin(2x)ex dx to both sides of the.
