Integral Calculus
Daniel Rakotonirina
February 17, 2017
1
Trigonometric Integrals
The goal of this section is to develop techniques for evaluating integrals of trigonometric functions.
Recall
• sin2 x + cos2 x = 1
• sin2 x =
1.1
1.1.1
1 − cos 2x
2
tan2 x + 1 = sec2 x
cos2 x =
1 + cos 2x
2
Integration of powers of signs
Powers of sin x or cos x
Z
Let m and n be positive integers. How to deal with integrals of the forms
Z
Rule for
sinm x dx or
Z
cosn x dx ?
sinm x dx
• m odd: Split off sinm into sinm−1 x · sin x, then rewrite the even power of sin x using the identity
sin2 x = 1 − cos2 x, then use substitution u = cos x and du = − sin x dx
Z
Z
Z
Z
m−1
m−1
x dx} = − (1 − u2 ) 2 du
sinm x dx = sinm−1 x · sin x dx = (1 − cos2 x) 2 |sin{z
−du
1 − cos 2x
• m even: Use half-angle formula sin2 x =
to transform the integration into polynomial
2
of cos 2x and then apply the substitution strategy for cos 2x
Z
Rule for
cosn x dx
• n odd: Split off cosn into cosn−1 x · cos x, then rewrite the even power of cos x using the identity
cos2 x = 1 − sin2 x, then use substitution u = sin x and du = cos x dx
Z
Z
Z
Z
n−1
n−1
cosn x dx = cosn−1 x · cos x dx = (1 − sin2 x) 2 cos
x
dx
=
(1 − u2 ) 2 du
| {z }
du
1 + cos 2x
• n even: Use half-angle formula cos2 x =
to transform the integration into polynomial
2
of cos 2x and then apply the substitution strategy for cos 2x
1
Z
Example 1.1. Evaluate the following integrals: I =
sin5 x dx
• Splitting off sin5 x = sin4 x · sin x
Z
Z
Z
I = sin4 x · sin x dx = (sin2 x)2 · sin x dx = (1 − cos2 x)2 · sin x dx
• Substitution u = cos x and du = − sin x dx
Z
Z
2u3
−u5
2 2
+
−u+c
I = (1 − u ) · (−du) = − (1 − 2u2 + u4 )du =
5
3
I =−
cos5 x 2 cos3 x
+
− cos x + c
5
3
Z
Example 1.2. Evaluate the following integrals: I =
cos4 x dx
1 + cos 2x
2
2
Z Z
Z
Z 1 + 2 cos 2x + cos2 2x
1 + cos 2x
4
2
2
dx =
I=
cos x dx = (cos x) dx =
dx
2
4
Z
Z
Z
Z
1
2 cos 2x
cos2 2x
x 1 1
1
I=
dx +
dx +
dx = + · sin 2x +
cos2 2x dx
4
4
4
4 2 2
4
{z
}
|
• half-angle formula cos2 x =
A
Z
1 + cos 4x
x 1
dx = + sin 4x + c
2
2 8
1 x 1
x 1
+ sin 4x + c
I = + sin 2x +
4 4
4 2 8
A=
1.1.2
Powers of sin x and cos x
Z
Here we are dealing with integrals of the form
Z
Rule for
sinm x cosn x dx
sinm x cosn x dx
• m odd & positive, n real: split off sin x as sinm x = sinm−1 x · sin x. Rewrite the even power of
sin x in terms of cos x using the identity sin2 x = 1 − cos2 x. Then use substitution u = cos x and
du = − sin x dx
Z
Z
Z
m−1
m
m−1
n
n
sin x cos x dx = sin
x · sin x · cos x dx = − (1 − u2 ) 2 un du
• n odd & positive, m real: split off cos x as cosn x = cosn−1 · cos x. Rewrite the even power of
cos x in terms of sin x using the identity cos2 x = 1 − sin2 x. The use substitution u = sin x and
du = cos x dx
Z
Z
Z
n−1
m
m
n
n−1
sin x cos x dx = sin x · cos
x · cos x dx = um (1 − u2 ) 2 du
1 − cos 2x
• m and n both even and non negative: use half-angle formulas sin2 x =
and
2
1 + cos 2x
cos2 x =
to transform the integral into polynomial function of cos 2x. Then apply the
2
previous strategy for the powers of cos 2x greater than 1
2
Z
Example 1.3. Evaluate the integral I =
Z
cos x · sin x dx =
I=
I=
Z 2
4
1
8
Z
cos4 x · sin2 x dx. Here we use half-angle, so:
1 + cos 2x
2
2
sin x dx =
(1 + 2 cos 2x + cos2 2x)(1 − cos 2x) dx =
x 1
I= +
8 8
Z 2
1
8
Z
1 + cos 2x
2
2 1 − cos 2x
2
2
dx
(1 + cos 2x − cos2 2x − cos3 2x) dx
Z
Z
1
1
1
cos2 2xdx −
cos3 2x dx
sin 2x −
2
8
8
Z
x sin 4x
+
+ c. So for the last integral, we get:
2
8
Z
Z
Z
Z
du
cos3 2x dx =
cos2 2x · cos 2x dx = (1 − sin2 2x) · cos 2x dx = (1 − u2 )
2
Z
Z
3
1
u3
sin 2x sin 2x
1
(1 − u2 )du =
cos3 2x dx =
u−
+c=
−
+c
2
2
3
2
6
We know from the previous example that
cos2 2x dx =
Therefore,
Z
I=
x 1
+
8 8
cos4 x · sin2 x dx =
Z
Example 1.4. Evaluate the integral I =
1
1 x sin 4x
1 sin 2x sin3 2x
sin 2x −
+
−
−
+C
2
8 2
8
8
2
6
cos5 x ·
√
Z
sin x dx =
1
cos5 x · sin 2 x dx.
• odd power for cos x. Split off cos x as cos5 x = cos4 x · cos x and use cos2 x = 1 − sin2 x. We have:
Z
Z
Z
√
1
1
5
2
2
2
I = cos x · sin x dx = (cos x) sin x · cos x dx = (1 − sin2 x)2 sin 2 x · cos x dx
• Substitution u = sin x and du = cos x dx. So we have:
Z
Z
Z
1
1
1
5
9
I = (1 − u2 )2 u 2 du = (1 − 2u2 + u4 )u 2 du = (u 2 − 2u 2 + u 2 )du
I=
2 3
u2 − 2
3
2 7
u2
7
+
3
7
11
2 11
2
4
2
u 2 + c = (sin x) 2 − (sin x) 2 + (sin x) 2 + c
11
3
7
11
Powers of tan x and sec x
1.1.3
Z
Here we are dealing with integrals of the form
tanm x secn x dx where m and n are positive integers.
• tan2 x = sec2 x − 1 and sec2 x = tan2 x + 1
•
d
d
tan x = sec2 x and
sec x = sec x · tan x
dx
dx
• u = tan x =⇒ du = sec2 x dx and u = sec x =⇒ du = sec x · tan x dx
Z
•
tan x dx = − ln | cos x| + c = ln | sec x| + c
Z
•
sec x dx = ln | sec x + tan x| + c
3
Z
Rule for
tanm x secn x dx
• n even: split off sec2 x as secn x = secn−2 x · sec2 x. Rewrite the remaining even power of
sec x, secn−2 x, in terms of tan x using identity sec2 x = tan2 x + 1. Then use substitution u = tan x
and du = sec2 x dx
Z
Z
Z
n−2
2
n−2
tanm x secn x dx = tanm x · sec
x
·
sec
x
dx
=
um (u2 + 1) 2 du
| {z }
(sec2 x)
|
{z
(tan2 +1)
n−2
2
}
n−2
2
• m odd: split off sec x · tan x as tanm x · secn x = tanm−1 x · secn−1 x · tan x · sec x. Rewrite the
remaining even power of tan x in terms of sec x using the identity tan2 x = sec2 x − 1. Then use
the substitution u = sec x and du = sec x · tan x dx
Z
Z
Z
m−1
m
n
m−1
n−1
x · tan x · sec x dx = (u2 − 1) 2 un−1 du
tan x sec x dx =
|tan {z x} · sec
(tan2 x)
{z
|
(sec2 x−1)
m−1
2
}
m−1
2
• m even and n odd: write the even power of tan x in terms of sec x to produce polynomial in
sec x use reduction formula (textbook p.526)
Z
Example 1.5. Evaluate the integral
Z
tan2 x sec4 x dx =
Z
tan2 x sec4 x dx
tan2 x sec2 x · sec2 x dx =
Z
tan2 x(1 + tan2 x) sec2 x dx
Now we let u = tan x and du = sec2 x dx. So we have,
Z
Z
u5
u3
tan2 x sec4 x dx = u2 (1 + u2 )du =
+
+c
3
5
Z
Example 1.6. Evaluate the following integral
Z
tan3 x sec7 x dx =
Z
tan3 x sec7 x dx
tan2 x sec6 x(tan x sec x) dx =
Z
(sec2 x − 1) sec6 x(tan x sec x) dx
Now we let u = sec x and du = sec x tan x dx. Therefore,
Z
Z
Z
u9
u7
sec9 x sec7 x
tan3 x sec7 x dx = (u2 − 1)u6 du = (u8 − u6 )du =
−
+c=
−
+c
9
7
9
7
Z
Example 1.7. Evaluate the following integral
Z
2
tan x sec x dx =
Z
tan2 x sec x dx
Z
2
(sec −1) sec x dx =
Then use the reduction formula in textbook p.527
4
3
sec x dx −
Z
sec x dx