MATH 1172 Midterm: MATH 1172 Ohio State University Math 1172 7.1 Solutions Fa 2014
31 Jan 2019
School
Department
Course
Professor
Math 1172 (Buenger) Section 7.1 September 15, 2015
Evaluate the following integrals:
1.
Zsin3x
cos5xdx
Solution:
Zsin3x
cos5xdx =Zsin3x
cos3x·1
cos2xdx
=Ztan3(x) sec2(x)dx
Let u= tan x. Then du = sec2x dx or
dx =du
sec2x. Thus
=Zu3sec2(x)du
sec2x
=Zu3du
=u4
4+C
=tan4x
4+C.
2.
Z1
x−1+ 1 dx
Solution:
Z1
x−1+ 1 dx =Z1
x−1+ 1 ·x
xdx
=Zx
1 + xdx
=Z1 + x−1
1 + xdx
=Z1 + x
1 + x−1
x+ 1 dx
=Z1−1
x+ 1 dx
=x−ln(x+ 1) + C.
3.
Z1
x−1+x−3dx
Solution:
Z1
x−1+x−3dx =Zx3
x2+ 1 dx.
Let u=x2+ 1. Then x2=u−1, and
du = 2x dx. Thus
Z1
x−1+x−3dx
=Zx2·x
u
du
2x
=1
2Zu−1
udu
=1
2Z1−1
udu
=1
2(u−ln |u|) + C
=1
2x2+ 1 −ln |x2+ 1|+C.
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MATH 1172 Full Course Notes
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Document Summary
Z sin3 x cos5 x dx = z sin3 x cos3 x. Then du = sec2 x dx or dx = du sec2 x. 1 x 1 + x 3 dx = z x3 x2 + 1 dx. Then x2 = u 1, and du = 2x dx. = x ln(x + 1) + c: r sin x+tan x cos2 x. Z sin x + tan x cos2 x dx sin x cos3 x dx cos2 x. Z sin x sin 2x dx = 2z sin x sin x cos x dx. Then du = sin x dx, and. Let u = sin x then du = cos x dx and dx. 2z sin2 x cos x dx = 2z u2 du cos2 x. Z sin x + tan x u2 + Z x x4 + 2x2 + 1 dx = z x (x2 + 1)2 dx. Then du = 2x dx or dx = du.
