Resolução de integrais trigonométricas

1. INTEGRATION BY SUBSTITUTION
Section 5.5
November 25, 2013

2. LAST TIME
By the FTC (Part I): If f is continuous on [a, b] and
F(x) =
x
a
f(t) dt =⇒ F (x) = f(x)
F(x) =
a
x
f(t) dt =⇒ F (x) = − f(x)
F(x) =
g(x)
a
f(t) dt =⇒ F (x) = f(g(x)) · g (x)

3. SUBSTITUTION
Substitution is a notational trick to make it easier to determine
indefinite integrals.
It uses something called a differential.
Definition
Let u = f(x) for some function f. The differential of u, denoted
du, is given by
du = f (x) dx.
For example,
u = 7x2
+ 1 =⇒ du = 14x dx
u = 24x
=⇒ du = 4 ln(2)24x
dx
For us, this is purely notational.

4. SUBSTITUTION
Determine (3x2
− 5)4
2x dx.
Let u = 3x2
− 5.
Then du = 6x dx =⇒ dx =
du
6x
Rewriting the above indefinite integral using u and du we get
(3x2
− 5)4
2x dx = u4
2x
du
6x
=
1
3
u4
du
This indefinite integral is easier to determine:
1
3
u4
du =
1
3
u4
du =
1
3
1
5
u5
+ C =
1
15
u5
+ C
Since u = 3x2
− 5, substituting back we get:
1
15
(3x2
− 5)5
+ C

5. SUBSTITUTION
What did we do?
(1) We chose a u and determined du.
(2) We rewrote our integral using u and du.
(3) Found the indefinite integral.
(4) Substitute back for u.
Question: How do we choose our u?
No foolproof way, but in general, choose:
the term under a root or raised to a power,
the exponent of an exponential function,
the logarithm,
the term in the denominator,
the term inside a trig function.

6. EXAMPLES
Determine x 1 − x2 dx.
u = 1 − x2
du = − 2x dx (dx = −du/2x)
Rewriting the above indefinite integral using u and du we get
x 1 − x2 dx = x
√
u
−1
2x
du = −
1
2
√
u du
This indefinite integral is easier to determine:
−
1
2
u1/2
du = −
1
2
1
1
2 + 1
u
1
2
+1
+ C = −
1
3
u3/2
+ C
Since u = 1 − x2, we get: −
1
3
(1 − x2
)3/2
+ C

7. EXAMPLES
Determine 3t2
e2t3
dt.
u = 2t3
du = 6t2 dt (dt = du/6t2)
Rewriting the above indefinite integral using u and du we get
3t2
e2t3
dt = 3t2
eu 1
6t2
du =
1
2
eu
du
This indefinite integral is easier to determine:
1
2
eu
du =
1
2
eu
+ C
Since u = 2t3, we get:
1
2
e2t3
+ C

8. EXAMPLES
Determine
1
x ln(x)
dx.
u = ln(x)
du =
1
x
dx (dx = xdu)
Rewriting the above indefinite integral using u and du we get
1
x ln(x)
dx =
1
xu
xdu =
1
u
du
This indefinite integral is easier to determine:
1
u
du = ln |u| + C
Since u = ln(x), we get: ln | ln(x)| + C

9. EXAMPLES
Determine
y2
2y3 + 1
dy.
u = 2y3 + 1
du = 6y2 dy (dy = du/6y2)
Rewriting the above indefinite integral using u and du we get
y2
2y3 + 1
dy =
y2
u
·
1
6y2
du =
1
6
1
u
du
This indefinite integral is easier to determine:
1
6
1
u
du =
1
6
ln |u| + C
Since u = 2y3 + 1, we get:
1
6
ln |2y3
+ 1| + C

10. EXAMPLES
Determine t sin(2t2
) dt.
u = 2t2
du = 4t dt (dt = du/4t)
Rewriting the above indefinite integral using u and du we get
t sin(2t2
) dt = t sin(u)
1
4t
du =
1
4
sin(u) du
This indefinite integral is easier to determine:
1
4
sin(u) du = −
1
4
cos(u) + C
Since u = 2t2, we get: −
1
4
cos(2t2
) + C

11. PRACTICE PROBLEMS
Use substitution to determine the following indefinite integrals:
(1) z(4z2
− 5)4
dz
(2) (1 − t)e2t−t2
dt
(3)
r
r2 + 2
dr
(4) θ2
sec2
(5θ3
+ 10) dθ

12. PRACTICE PROBLEMS
(1) z(4z2
− 5)4
dz.
u = 4z2 − 5
du = 8z dz (dz = du/8z)
Rewriting the above indefinite integral using u and du we get
z(4z2
− 5)4
dz = zu4 1
8z
du =
1
8
u4
du
This indefinite integral is easier to determine:
1
8
u4
du =
1
40
u5
+ C
Since u = 4z2 − 5, we get:
1
40
(4z2
− 5)5
+ C

13. PRACTICE PROBLEMS
(2) (1 − t)e2t−t2
dt.
u = 2t − t2
du = (2 − 2t) dt (dt = du/(2 − 2t))
Rewriting the above indefinite integral using u and du we get
(1 − t)e2t−t2
dt = (1 − t)eu 1
2 − 2t
du =
1
2
eu
du
This indefinite integral is easier to determine:
1
2
eu
du =
1
2
eu
+ C
Since u = 2t − t2, we get:
1
2
e2t−t2
+ C

14. PRACTICE PROBLEMS
(3)
r
r2 + 2
dr.
u = r2 + 2
du = 2r dr (dr = du/2r)
Rewriting the above indefinite integral using u and du we get
r
r2 + 2
dr =
r
u
1
2r
du =
1
2
·
1
u
du
This indefinite integral is easier to determine:
1
2
1
u
du =
1
2
ln |u| + C
Since u = r2 + 2, we get:
1
2
ln |r2
+ 2| + C

15. PRACTICE PROBLEMS
(4) θ2
sec2
(5θ3
+ 10) dθ.
u = 5θ3 + 10
du = 15θ2 dθ (dθ = du/15θ2)
Rewriting the above indefinite integral using u and du we get
θ2
sec2
(5θ3
+ 10) dθ = θ2
sec2
(u)
1
15θ2
du =
1
15
sec2
(u) du
This indefinite integral is easier to determine:
1
15
sec2
(u) du =
1
15
tan(u) + C
Since u = 5θ3 + 10, we get:
1
15
tan(5θ3
+ 10) + C

16. SUBSTITUTION
It may be the case that after substituting for u and du the original
variable remains in the integrand.
Determine x(5x − 1)2/3
dx
u = 5x − 1
du = 5 dx (dx = du/5)
x(5x − 1)2/3
dx =
1
5
xu2/3
du
What do we do about the x? Try to rewrite it in terms of u,
u = 5x − 1 ⇒ x = (u + 1)/5.
1
5
xu2/3
du =
1
25
(u + 1)u2/3
du =
1
25
u5/3
+ u2/3
du
3
200
u8/3
+
3
125
u5/3
+ C =
3
200
(5x − 1)8/3
+
3
125
(5x − 1)5/3
+ C

17. EXAMPLE
Determine p3
(p2
+ 1)5
dp.
u = p2 + 1
du = 2p dp (dp = du/2p)
Rewriting the above indefinite integral using u and du we get
p3
2p
u5
du =
1
2
p2
u5
du
Since u = p2 + 1, we know p2 = u − 1
1
2
p2
u5
du =
1
2
(u − 1)u5
du =
1
2
u6
− u5
du =
1
14
u7
−
1
12
u6
+ C
Since u = p2 + 1, we get:
1
14
(p2
+ 1)7
−
1
12
(p2
+ 1)6
+ C