Download 8.1: 8 8.1: 68 - UC Davis Mathematics

8.1: 8
Evaluate the following integral using integration by parts:
Z
xe3x dx.
We choose u to be x. Then we have:
dv = e3x dx
1
v = e3x
3
u=x
du = dx
Applying the integration by parts formula gives:
Z
Z
1
1 3x
1
1
xe3x dx = xe3x −
e = xe3x − e3x + C.
3
3
3
9
8.1: 68
Use the fomula
Z
f −1 (x) dx = xf −1 (x) −
Z
f (y) dy,
(1)
where y = f −1 (x), to evaluate the integral:
Z
tan−1 (x) dx.
Express your answer in terms of x.
By the formula, we have:
Z
tan−1 (x) dx = x tan−1 (x) −
Z
tan y dy
= x tan−1 (x) − ln(cos y) + C
= x tan−1 (x) − ln(cos(tan−1 x)) + C
1
= x tan−1 (x) + ln 1 + x2 + C
2
There are two steps in there that might require some explaining. To integrate tan y, rewrite tan y as
sin y/ cos y, and make a u-substitution. And to simplify cos(tan−1 x), let θ = tan−1 x, and draw a reference
triangle with the side opposite of θ having length x and the side adjacent to θ having length 1. Taking the
cos of θ should then give you something similar to my result, though you’ll have to use ln rules to manipulate
it to the form I’ve given.
1
8.2: 24
Evaluate the integral:
π
Z
√
1 − cos 2x dx
0
To evaluate this integral we’ll need to use the trig identity:
1 − cos 2θ
.
2
Rearranging this gives something that looks quite similar to our integrand:
√
√
2 sin θ = 1 − cos 2θ.
sin2 θ =
Plugging this in, we see that:
Z π
√
1 − cos 2x dx =
√ Z
2
0
0
π
π
√
√
sin x dx = − 2 cos = 2 2.
0
8.2: 34
Evalute the integral:
Z
sec x tan2 x dx.
Sometimes a small change can make a big difference - look at problem 33, which is ever so slightly different
but can be done with a simple u-substitution of u = sec x. This one is not so simple. One way to go is to
integrate by parts. Let:
u = tan x
dv = sec x tan x dx
2
du = sec x dx
v = sec x
Then applying the integration by parts formula, we find that:
Z
Z
sec x tan2 x dx = sec x tan x − sec3 x dx.
Here we use the fact that sec2 x = 1 + tan2 x. Then:
Z
Z
Z
sec x tan2 x dx = sec x tan x − sec x dx − sec x tan2 x dx.
Notice now that the final integral on the right is what we started with. We can move this back over to the
left to get:
Z
Z
2 sec x tan2 x dx = sec x tan x − sec x dx
All that remains is to integrate sec x. To do this, multiply by 1 as follows:
Z
Z
Z
sec2 x + sec x tan x
sec x + tan x
dx =
dx.
sec x dx = sec x
sec x + tan x
sec x + tan x
Notice now that the numerator is the derivative of the denominator, and so we can apply the u-substitution
u = sec x + tan x. Putting everything together gives:
Z
1
1
sec x tan2 x dx = sec x tan x − ln (sec x + tan x) .
2
2
2
8.3: 4
Evaluate the integral:
2
Z
1
dx.
8 + 2x2
0
We first put this into a better form for working with:
Z 2
Z
1
1 2
1
dx =
dx.
2
x 2
8
+
2x
8
0
0 1+
2
x
2
We now make the substitution
our limits!), we get:
1
8
Z
2
0
1
1+
= tan θ. Then dx = 2 sec2 θ dθ. Plugging in (and not forgetting to change
dx =
x 2
2
1
8
Z
tan−1 (1)
tan−1 (0)
1
1
2 sec2 θ dθ =
4
1 + tan2 θ
Z
π/4
dθ =
0
π
.
16
8.3: 38
Use an appropriate substitution and then a trigonometric substitution to evaluate the integral:
Z e
1
p
dy.
2
1 y 1 + (ln y)
We begin by letting x = ln y, which gives dx =
e
Z
1
1
y
dy. Then substituting gives:
Z
1
y
p
1 + (ln y)2
dy =
1
√
0
1
dx.
1 + x2
We now let x = tan θ, and dx = sec2 θ dθ. Then substitution gives:
1
Z
√
0
1
dx =
1 + x2
Z
π/4
√
0
1
1 + tan2 θ
sec2 θ dθ.
Simplifying gives an integral we’ve seen before in these solutions:
Z
0
π/4
√
1
1 + tan2 θ
2
Z
sec θ dθ =
0
π/4
π/4
√
sec θ dθ = ln(sec θ + tan θ)
= ln(1 + 2).
0
3