# 2009 Test 3 solution

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School
Department
Course
Professor MAT137Y 2009–2010 Winter Session, Solutions to Term Test 3
1. Evaluate the following integrals.
(8%) (i) Zdx
x(lnx)5/2.
Let u=lnx. Then du =1
xdx, thereby giving us
Zdu
u5/2=Zu5/2du =2
3u3/2+C=2
3(lnx)3/2+C.
(8%) (ii) Z1
0
arctanx dx.
We integrate by parts. Let u=arctanx,dv =dx. Then du =dx/(1+x2)and v=x. This yields,
Z1
0
arctanx dx =hxarctanxi1
0Z1
0
x
1+x2dx =xarctanx1
2ln(1+x2)1
0
=arctan1 1
2ln2 0=π
41
2ln2.
(10%) (iii) Zdx
(2+x2)2.
We apply a trig substitution: let x=2 tan θ. Then dx =2 sec2θdθ. This gives
Z2sec2θ
(2+2tan2θ)2dθ=Z2sec2θ
4sec4θdθ=2
4Zcos2θdθ.
Using trig identities,
2
4Zcos2θdθ=2
41
2θ+1
4sin2θ+C=2
8θ+2
8sinθcosθ+C.
We substitute back in terms of x. We draw a right triangle with xbeing the length of the opposite
side and 2 being the length of the adjacent side. By the Pythagorean theorem the length of the
hypoteneuse is x2+2. Therefore sinθ=x
x2+2and cosθ=2
x2+2. Hence,
Zdx
(2+x2)2=2
8arctanx
2+2
8
x
x2+2
2
x2+2+C
=2
8arctanx
2+1
4x
x2+2+C.
(10%) (iv) Z1x+2x2x3
x(x2+1)2dx.
Let Ibe the integral above. We re-write the integral using partial fractions:
I=ZA
x+Bx +C
x2+1+Dx +E
(x2+1)2dx.
1
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