Download Week 13 – Trigonometric Integrals

Week 13 – Trigonometric Integrals
Powers of Sines and Cosines
To evaluate trigonometric integrals of the type
Z
sinn x cosm x dx
• If n is odd, use the substitution:
u = cos x,
du = − sin x dx
and then the identity cos2 x + sin2 x = 1
if m is odd, try: u = sin x,
du = cos x dx
If both are odd, it doesn’t matter.
• If both n and m are even, use the identities
sin2 x =
1 − cos 2x
2
1 + cos 2x
2
to simplify the integral.
cos2 x =
Z
sin3 x cos2 x dx
Exercise 13-1: Evaluate
Solution: Using u = cos x, du = − sin x dx we
obtain:
Z
3
Z
2
sin2 x cos2 x sin x dx
sin x cos x dx =
Z
=
(1 − u2 )u2 (−du)
Z
=
u4 − u2 du
=
u5 u3
−
+c
5
3
=
cos5 x cos3 x
−
+c
5
3
Z
Exercise 13-2: Evaluate
sin2 x cos2 x dx
Solution:
Z
1 − cos 2x 1 + cos 2x
·
dx
2
2
1
=
4
Z
1
=
4
Z
1
=
8
Z
=
1 − cos2 2x dx
1−
1 + cos 4x
dx
2
1 − cos 4x dx
x sin 4x
−
+c
8
32
Z
Exercise 13-3: Evaluate
cos7 x dx
Solution:
Z
Z
7
cos x dx =
cos6 x · cos x dx
Z
=
Z
=
Z
=
(1 − sin2 x)3 · cos x dx
(1 − u2 )3 du
u = sin x
(1 − 3u2 + 3u4 − u6 ) du
3
1
= u − u3 + u5 − u7 + c
5
7
3
1
= sin x − sin3 x + sin5 x − sin7 x + c
5
7
Exercise 13-4: Evaluate the following integrals:
Z
a) sin6 x cos3 x dx
Z
b)
Z
c)
Z
d)
Z
e)
sin5 x cos5 x dx
sin8 x cos x dx
cos4 x dx
sin4 x cos4 x dx
Products of Sines and Cosines
Adding the identities
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
we obtain:
sin(A + B) + sin(A − B) = 2 sin A cos B
In other words
1
sin(A + B) + sin(A − B)
2
Similarly, starting with
sin A cos B =
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
we obtain:
1
cos A cos B =
cos(A + B) + cos(A − B)
2
1
sin A sin B =
cos(A − B) − cos(A + B)
2
Z
Exercise 13-5: Evaluate
sin 7x cos 8x dx
Solution: Using the above identities, we obtain:
Z
Z
1
sin 7x cos 8x dx =
sin 15x − sin x dx
2
= −
cos 15x cos x
+
+c
30
2
Exercise 13-6: Evaluate the following integrals:
Z
a) sin 5x sin 3x dx
Z
b)
sin 4x cos 8x dx
Z
c)
cos 3x cos
x
dx
2
Integrals containing secant and
tangent
Remember the formulas:
d
tan x = sec2 x
dx
d
sec x = sec x tan x
dx
and
sec2 x = 1 + tan2 x
Z
Exercise 13-7: Evaluate
sec x dx
Solution: Using the derivative formulas for sec x
and tan x, we obtain:
d
(sec x + tan x) = sec x (sec x + tan x)
dx
Z
Z
sec x dx =
sec x (sec x + tan x)
dx
sec x + tan x
The substitution u = sec x + tan x gives:
Z
=
du
u
= ln |u| + c
= ln | sec x + tan x| + c
Z
Exercise 13-8: Evaluate
Solution:
Z
Z
2
tan x dx =
tan2 x dx
sec2 x − 1 dx
= tan x − x + c
Z
Exercise 13-9: Evaluate tan3 x dx
Solution:
Z
Z
3
tan x dx =
Z
=
tan2 x · tan x dx
(sec2 x − 1) · tan x dx
Z
Z
2
sec x · tan x dx −
=
Z
Z
u du −
=
tan x dx
tan x dx
where u = tan x
=
tan2 x
− ln | sec x| + c
2
Exercise 13-10: Evaluate the following integrals:
Z
a) tan4 x dx
Z
b)
Z
c)
Z
d)
tan3 x sec2 x dx
tan3 x sec x dx
tan2 x sec4 x dx
Trigonometric Substitutions
Using certain trigonometric identities, we can
simplify integrals involving square roots:
√
a2 − x2 , use x = a sin θ to obtain:
π
π
a| cos θ|, − 6 θ 6
2
2
√
• For a2 + x2 , use x = a tan θ to obtain:
π
π
a| sec θ|, − < θ <
2
2
√
• For x2 − a2 , use x = a sec θ to obtain:
π
a| tan θ|, 0 6 θ <
if x > a
2
π
< θ 6 π if x > −a
2
• For
Exercise 13-11: Evaluate the following integrals:
Z
dx
√
a)
1 − x2
Z
dx
√
b)
9 − x2
Z √
c)
4 − x2 dx
Z
d)
Z
e)
Z
f)
Z
g)
√
√
√
x dx
4 − x2
x dx
4 + x2
dx
9 + x2
√
dx
x2 − 16
Z
h)
√
x2 dx
16 − x2
Review Exercises
Exercise 13-12: Evaluate the following integrals:
Z
a) cos6 x dx
Z
b)
cos6 x sin3 x dx
Z
c)
cos(πx) sin(2πx) dx
Z
d)
Z
e)
Z
f)
Z
g)
Z
h)
Z
i)
tan8 x sec4 x dx
tan3 x sec5 x dx
sec3 x dx
√
√
x2 dx
25 − x2
dx
9x2 − 16
Z
dx
√
x 2 x2 − 4
Z
dx
√
x2 x2 + 1
j)
k)
cos3 (πx) dx
— End of WEEK —
Author: Dr. Emre Sermutlu
Last Update: December 20, 2016