MATH 1A SECTION: NOVEMBER 20, 2013
Compute the following integrals!
1.
6.
Z
2
1
Z
3
(3t − 1)50 dt
x cos(x ) dx.
0
7.
2.
Z
ex sin(ex ) dx.
Z
x ln x dx.
8.
1/2
Z
3.
Z
dx
.
5 − 3x
0
9.
arcsin x
√
dx.
1 − x2
Z
cot x dx.
4.
Z
1
(x2 + 1)e−x dx.
10.
0
Z
e2θ sin 3θ dθ.
5.
Z
ln t dt.
11.
Z
π/3
−π/3
x4 sin x dx.
Solutions and Commentary
1. Here, we use integration by substitution. Substitute u = x3 . Then du = 3x2 dx, and
hence 31 du = x2 dx, so therefore we have
Z
Z
1
1
1
2
3
cos u du = sin u + C = sin(x3 ) + C .
x cos(x ) dx =
3
3
3
It’s important to write our final answer in terms of x and not u; at the end, we undid
our substitution. Also, remember to include the constant of integration when evaluating
indefinite integrals.
General advice for integration by substitution: Once you decide that you want to integrate by substitution, carefully write down everything, and all at once, convert everything
to your new variable. Here, we started with a dx integral and converted directly into a
du integral; we did not mix x and u during the computation.
2. This problem requires integration by parts. Recall that this is
Z
Z
u dv = uv − v du.
In general, we want to take the integrand and find a piece that is nice when differentiated
(call this u) and a piece that is nice when integrated (call this dv). This is sometimes
tricky – there sometimes isn’t an obvious way to select u and dv, and you may need to
experiment Ra bit.
We have x ln x dx, so we set u = ln x and dv = x dx. Note here that we know how
to differentiate ln x but not how to integrate it, so therefore it makes sense for ln x to be
part of the u and not the dv. Also, note that we’ve included the dx as part of the dv
piece. Then, we have
u = ln x
v = 21 x2
1
du = x dx dv = x dx.
It remains to simply look everything up in our table. We have
Z
Z
Z
x ln x dx = u dv = uv − v du
1
= x2 ln x −
2
Z
1
x
1 2
x
2
1
dx = x2 ln x −
2
Z
x
1
x2
dx = x2 ln x −
+C .
2
2
4
3. Substitute u = 5 − 3x. Then du = −3 dx, so dx = − 31 du. Therefore,
Z
Z
Z
dx
1
1
1
1
1
1
=
− du = −
du = − ln u + C = − ln(5 − 3x) + C .
5 − 3x
u
3
3
u
3
3
R1
4. We have 0 (x2 +1)e−x dx, and we integrate by parts. We want to make the x2 +1 piece go
away since we know how to integrate e−x , so that’s the piece that we want to differentiate.
So that gives
u = x2 + 1 v = −e−x
du = 2x dx dv = e−x dx.
2
So plugging this in gives
Z 1
Z 1
Z 1
2
−x
1
(x + 1)e dx =
u dv = [uv]0 −
v du
0
0
0
Z 1
Z
2
−x 1
−x
−1
= −(x + 1)e 0 +
2xe dx = −2e + 1 + 2
0
1
xe−x dx.
0
At this point, what do we do with the remaining integral? We should integrate by parts
again! So that gives
u=x
v = −e−x
du = 1 dx dv = e−x dx.
Hence
Z 1
Z 1
2
−x
−1
(x + 1)e dx = −2e + 1 + 2
xe−x dx
0
0
Z 1
−1
−x 1
−x
= −2e + 1 + 2 −xe 0 +
e dx
0
1 = −2e−1 + 1 + 2 −e−1 + −e−x 0
= −2e−1 + 1 + 2 −e−1 − e−1 + 1 = 3 − 6e−1 .
5. This is a bit tricky! This doesn’t look like a product of two things, but it actually is
integration by parts. If you like, think of one of the two things in the product as being 1.
That is, u = ln t and dv = 1 dt = dt. Then we have
u = ln t
v=t
1
du = t dt dv = dt.
Then integration by parts yields
Z
Z
Z
Z
Z
1
ln t dt = u dv = uv − v du = t ln t − t · dt = t ln t − 1 dt = t ln t − t + C .
t
6. Here, substitute u = 3t − 1. Then du = 3 dt and hence 13 du = dt. Also, when t = 0, we
have u = −1; when t = 1, we have u = 2. This gives:
2
Z 1
Z
1 2
1 u51
1 251
1
251 + 1
50
50
(3t − 1) dt =
u du =
=
+
=
.
3 u=−1
3 51 −1 3 51
51
153
t=0
Note that it is really important to also update the bounds of integration when we do
substitution in a definite integral. Here, I explicitly wrote down t = . . . and u = . . . in
the bounds, in order to remind ourselves to do that; I recommend make a habit of doing
this to avoid confusion.
7. We substitute u = ex ; then du = ex dx. So our integral is
Z
Z
x
x
e sin(e ) dx = sin u du = − cos u + C = − cos(ex ) + C .
Note: If you can guess the antiderivative (say by doing substitution in your head),
that’s ok too. But be sure you don’t make a mistake! There’s no reason ever to compute
3
an antiderivative incorrectly; you can always differentiate your antiderivative and check
that it matches your integrand!
1
π
1
8. Set u = arcsin x. Then du = √1−x
2 dx. Also, arcsin 0 = 0 and arcsin 2 = 6 . So we have
Z π/6
Z 1/2
π/6 1 π 2
arcsin x
π2
√
u du = f rac12u2 0 =
dx =
.
=
2 6
72
1 − x2
u=0
x=0
9. Write cot x =
cos x
.
sin x
Then we want to compute
Z
Z
cos x
cot x dx =
dx.
sin x
Now, it might be easier to see what we should do. We substitute u = sin x; then du =
cos x dx, so hence
Z
Z
Z
cos x
1
cot x dx =
dx =
du = ln u + C = ln(sin x) + C .
sin x
u
10. This is tricky. We integrate twice by parts, and then solve for what we want.
u = e2θ
v = − 31 cos 3θ
2θ
du = 2e dθ dv = sin 3θ dθ.
Then
Z
Z
1 2θ
2
2θ
e sin 3θ dθ = − e cos 3θ +
e2θ cos 3θ dθ.
3
3
We integrate by parts again, this time with
u = e2θ
v = 31 sin 3θ
du = 2e2θ dθ dv = cos 3θ dθ.
This gives
Z
Z
1 2θ
2
2θ
e2θ cos 3θ dθ
e sin 3θ dθ = − e cos 3θ +
3
3
Z
1 2θ
2 1 2θ
2
2θ
= − e cos 3θ +
e sin 3θ −
e sin 3θ dθ
3
3 3
3
Z
1 2θ
2 2θ
4
= − e cos 3θ + e sin 3θ −
e2θ sin 3θ dθ.
3
9
9
It might seem like we haven’t done anything here; we still have the same integral that we
wanted to compute. But we can now solve for it! Rearrange to get
Z
13
1
2
e2θ sin 3θ dθ = − e2θ cos 3θ + e2θ sin 3θ,
9
3
9
so therefore
Z
9
1 2θ
2 2θ
9
1 2θ
2 2θ
2θ
e sin 3θ dθ =
− e cos 3θ + e sin 3θ + C . =
− e cos 3θ + e sin 3θ + C .
13
3
9
13
3
9
4
11. There’s a lot of ways to do this. One is to integrate by parts four times, so as to differentiate the x4 piece four times to make it go away. This is the same spirit of what we did
in problem 4, but messier. I won’t write out this computation.
More simply, we notice that we are integrating an odd function (symmetric about the
origin) over a symmetric interval around zero. If you imagine plotting this, our function
has equal area above and below the x-axis, so those areas should cancel out and our
integral should be zero.
To formalize this intuition, we can also evaluate this integral via substitution. Let
u = −x, so u4 = x4 and du = −dx. Then
Z π/3
Z −π/3
Z π/3
4
4
x sin x dx =
−u sin u (−du) = −
u4 sin u du.
x=−π/3
u=π/3
u=−π/3
Here, we keep track of our minus signs carefully; note that we’ve used the fact that reverse
the bounds of integration will also reverse the sign of the integral. Now we have the two
sides of the equation are negatives of each other, so
Z π/3
Z π/3
4
x4 sin x dx.
x sin x dx = −
−π/3
−π/3
Rearranging this gives
Z
π/3
x4 sin x dx = 0
2
−π/3
and hence
Z
π/3
x4 sin x dx = 0 .
−π/3
This problem was unusually tricky.
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