M408S Quiz 4 Solutions
1.
Sep 22, 2016
(5 pts)πFirst make a substitution and then use integration by parts to evaluate
´ 2 cos t
1
sin(2t)dt. You may need a double angle formula.
2 0 e
Solution 1.
1
2
1
2
ˆ
=
ˆ
π
2
ecos t sin(2t)dt
0
ˆ
π
2
ecos t 2 sin(t) cos(t)dt
0
π
2
=
ecos t sin(t) cos(t)dt
0
ˆ
π
2
=
ecos t cos(t) sin(t)dt
=
(let u = cost, du = −sintdt)
ˆ 0
eu u(−1)du
0
1
(change variable name for integration by parts)
ˆ
1
ex xdx
=
0
=
(let u = x, dv = ex dx, then du = 1dx, v = ex )
ˆ 1
x=1
xex |x=0 −
1 · ex dx
=
1e1 − 0e0 − ex |x=0
0
Solution 2.
x=1
=
e − 0 − (e − 1)
=
1
Use indenite integral: with the same substitution above,
1
2
ˆ
ˆ
ecos t sin(2t)dt
ecos t cos(t) sin(t)dt
=
ˆ
eu u(−1)du
ˆ
= − ex xdx
ˆ
= −(xex − 1ex dx)
=
= −(xex − ex )
(plug back x = u = cost
cost
= −(coste
= e
cost
−e
(1 − cost)
cost
)
after
you nd the antiderivative)
M408S Quiz 4 Solutions
Sep 22, 2016
t= π
Then plug in the bounds: ecost (1 − cost)|t=02 = e0 (1 − 0) − e1 (1 − 1) = 1.
Solution 3. Use integration by parts after using sin(2t) = 2sintcost:
I
1
2
ˆ
=
ˆ
ecos t sin(2t)dt
ecos t cos(t) sin(t)dt
=
Let u = cost, dv = ecost sintdt.
Because you can observe (ecost )0 = ecost (−sint).
Then du = −sintdt, v = −ecost .
ˆ
=
udv
ˆ
= uv − vdu
ˆ
= cost(−ecost ) − (−ecost )(−sint)dt
ˆ
cost
= −coste
− ecost sintdt
(do substitution u = cost, then sintdt = −du)
ˆ
cost
= −coste
− eu (−1)du
= −costecost + eu
= −costecost + ecost
Then plug in the bounds same as in Solution 2.
2.
´
(5 pts) Compute tan3 xdx. Given the formula:
* 1 point bonus: compute
Solution 1.
´
´
tanxdx = −ln(|cosx|) + C = ln(|secx|) + C.
´
tan5 xdx. It will reduce to the case of tan3 xdx.
ˆ
tan3 xdx
ˆ
tan2 xtanxdx
=
ˆ
(sec2 x − 1)tanxdx
ˆ
ˆ
=
sec2 xtanxdx − tanxdx
=
(do substitution u = tanx, then du = sec2 xdx)
ˆ
=
udu − (−ln|cosx|)
=
u2
+ ln|cosx|
2
M408S Quiz 4 Solutions
=
or :
Solution 2.
=
Sep 22, 2016
tan2 x
+ ln|cosx| + C
2
tan2 x
− ln|secx| + C
2
Similar as above,
ˆ
tan3 xdx
ˆ
ˆ
=
ˆ
sec2 xtanxdx −
=
tanxdx
ˆ
secx · secxtanxdx − tanxdx
(do substitution u = secx, then du = tanxsecxdx)
ˆ
=
udu − (−ln|cosx|)
=
sec2 x
+ ln|cosx| + C
2
The solution is same as tan2 x + ln|cosx| + C because tan2 x and sec2 x dier
by a constant ( 21 ).
sinx
Note. You may have lots of other ways to do this problem using tanx =
cosx
1
and secx = cosx .
Bonus. One of the solutions is:
2
2
2
ˆ
tan5 xdx
ˆ
tan3 x(sec2 x − 1)dx
ˆ
ˆ
=
tan3 xsec2 xdx − tan3 xdx
=
(observe that we already have
ˆ
tan3 dx from the rst part)
(do substitution u = tanx, du = sec2 xdx)
ˆ
tan2 x
=
u3 du − (
+ ln|cosx|)
2
tan4 x tan2 x
=
−
− ln|cosx| + C
4
2
You can also set tan5 x = tanx(sec2 x − 1)2 = tanx(sec4 x ´− 2sec2 x + 1) =
sec4 xtanx − 2sec2 xtanx + tanx and use U-substitution for sec4 xtanx and
´
4
sec2 xtanx. You end up with sec4 x − tan2 x − ln|cosx| + C , which is equivalent
to the above solution (by plugging in sec2 x = tan2 x + 1).