Answer Key for Quiz 1 (section A)
1. If we substitute u=x4+x2+ 1 then du = (4x3+ 2x)dx = 2x(2x2+ 1)dx, so du
2= (2x2+ 1)x dx and we
have
Z(x4+x2+ 1)7(2x2+ 1)x dx =Zu7du
2=1
2Zu7du
=1
2
u8
8+C=1
16(x4+x2+ 1)8+C.
The substitution v=x2isn’t powerful enough to knock off the original integral, but it does simplify it,
which is always worthwhile: dv = 2x dx, so dv
2=x dx and we have
Z(x4+x2+ 1)7(2x2+ 1)x dx =1
2Z(v2+v+ 1)7(2v+ 1)dv.
Now letting w=v2+v+ 1, in which case dw = (2v+ 1)dv, gives us
Z(x4+x2+ 1)7(2x2+ 1)x dx =1
2Z(v2+v+ 1)7(2v+ 1)dv
=1
2Zw7dw =w8
16 +C=1
16(x4+x2+ 1)8+C
as before.
2. (i) Because sin2θ+ cos2θ= 1, we can write
Zdθ
sin θcos θ=Z1·dθ
sin θcos θ=Z(sin2θ+ cos2θ)dθ
sin θcos θ.
Then we can split this up and simplify:
Z(sin2θ+ cos2θ)dθ
sin θcos θ=Zsin2θ dθ
sin θcos θ+Zcos2θ dθ
sin θcos θ
=Zsin θ dθ
cos θ+Zcos θ dθ
sin θ.
(ii) We can do these last two integrals by substituting u= cos θin the first one, and v= sin θin the second
one. We have du =−sin θ dθ and dv = cos θ dθ respectively, so
Zsin θ dθ
cos θ=Z−du
u=−ln |u|+C=−ln |cos θ|+C
and Zcos θ dθ
sin θ=Zdv
v= ln |v|+C= ln |sin θ|+C.
Therefore Zdθ
sin θcos θ= ln |sin θ|+C−ln |cos θ|+C,
and if we want to we can also write this as
Zdθ
sin θcos θ= ln |tan θ|+C.
There are several other ways to do the integral, of which I’ll show you only one:
Zdθ
sin θcos θ=Zdθ
sin θ
cos θcos2θ=Zsec2θ dθ
tan θ
where sec θis the reciprocal of cos θ. (One would not learn from our textbook that there is such a function
as sec θ, but some of you may have picked it up on the street somewhere.) If we now let w= tan θ, then
dw = sec2θ dθ and we have
Zdθ
sin θcos θ=Zsec2θ dθ
tan θ=Zdw
w= ln |w|+C= ln |tan θ|+C
as before.
