2006 Test 3 solution

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10 Apr 2012
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MAT 137Y 2006-2007 Winter Session, Solutions to Term Test 3
1. Evaluate the following integrals.
(8%) (i) Zsec3xtanx dx.
Zsec3xtanx dx =Zsec2xsecxtanx dx =1
3sec3x+C(by a simple substitution u=secx).
(8%) (ii) Z1
0
x·3xdx.
Integrating by parts, let u=x,dv =3xdx. Then du =dx and v=3x
ln3 . Hence
Z1
0
x·3xdx =x3x
ln3
1
0Z1
0
3x
ln3 dx =x3x
ln3 3x
(ln3)21
0
=3
ln3 3
(ln3)20+1
(ln3)2
=3
ln3 2
(ln3)2=3ln3 2
(ln3)2.
(10%) (iii) Zdx
x4x21.
Let x=secθ. Then dx =sec θtan θdθ, thus we have
Zsecθtanθ
sec4θsec2θ1dθ=Ztanθ
sec3θtanθdθ=Zdθ
sec3θ=Zcos3θdθ.
To integrate cos3θwe take out one power of cosθand apply a substitution:
Zcos3θdθ=Z(1sin2θ)cosθdθ=Zcosθsin2θcosθdθ=sinθ1
3sin3θ+C.
Substituting back, we can use triangles to see that if sec θ=x
1, then sin θ= (x21)/x. Hence
Zdx
x4x21=x21
x1
3 (x21)3/2
x3!+C.
(10%) (iv) Zdx
x3(x+1)2.
We decompose the integrand using partial fractions:
1
x3(x+1)2=A
x+B
x2+C
x3+D
x+1+E
(x+1)2
Obtaining a common denominator on the right side (and ignoring the denominator) gives
1=Ax2(x+1)2+Bx(x+1)2+C(x+1)2+Dx3(x+1) + Ex3
=A(x4+2x3+x2) + B(x3+2x2+x) +C(x2+2x+1) + D(x4+x3) + Ex3
= (A+D)x4+ (2A+B+D+E)x3+ (A+2B+C)x2+ (B+2C)x+C
Matching coefficients gives us C=1, B=2, A=3, D=3, and E=1. Hence
Z3
x2
x2+1
x33
x+11
(x+1)2dx =3ln|x|+2
x1
2x23ln|x+1|+1
x+1+C.
1
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