Download Ch 7.2 Trig Integrals

Math 133
Trigonometric Integrals
Stewart §7.2
Products by substitution. In this section, we develop several methods to find indefinite integrals (antiderivatives) of products of trig functions. The simplest method is a
simple trig substitution which reduces the integral to a polynomial:
(a)
R
sinn (x) cos2m+1 (x) dx for m ≥ 0. Take u = sin(x), du = cos(x) dx. Example:∗
R
R
sin−10 (x) cos3 (x) dx =
sin−10 (x) cos2 (x) · cos(x) dx
R
=
sin−10 (x)(1 − sin2 (x)) · cos(x) dx
R −10
=
u (1 − u2 ) du
R −10
=
u
− u−8 du
=
=
=
(b)
R
sin2n+1 (x) cosm (x) dx for n ≥ 0. Take u = cos(x), du = − sin(x) dx. Example:
R
R
sin5 (x) cos20 (x) dx = − sin4 (x) cos20 (x) · (− sin(x)) dx
R
= − (1 − cos2 (x))2 cos20 (x) · (− sin(x)) dx
R
=
(1 − u2 )2 u20 du
R
=
(1 − 2u + u2 )u20 du
=
=
(c)
R
1 21
2 22
1 23
21 u − 22 u + 23 u
1
1
21
22
21 cos (x) − 11 cos (x)
=
∗
†
R
+
1
23
cos23 (x).
tann (x) sec2m+2 (x) dx for m ≥ 0. Take u = tan(x), du = sec2 (x) dx. Example:†
R
R
tan(x)−1 sec6 (x) dx =
tan(x)−1 sec4 (x) · sec2 (x) dx
R
=
tan(x)−1 (tan2 (x) + 1)2 du
R −1 2
=
u (u + 1)2 du
R −1 4
=
u (u + 2u2 + 1) du
=
(d)
1 −7
1 −9
− −7
u
−9 u
− 91 sin−9 (x) + 17 sin−7 (x)
− 91 csc9 (x) + 17 csc7 (x).
1 4
2
4 u + u + ln|u|
1
4
2
4 tan (x) + tan (x)
+ ln|tan(x)|.
tan2n+1 (x) secm (x) dx for n ≥ 0. Take u = sec(x), du = tan(x) sec(x).
For brevity, we again neglect the +C in indefinite integrals, though you should write it on a test.
1
Notation: tan(x)−1 = tan(x)
= cot(x), but tan−1 (x) = arctan(x).
Remaining cases. If a product of trig functions does not fit the above types, we have
some strategies to make it tractable.
• Rewrite using trig function definitions, producing a good type.
Z
Z
Z
cos4 (x)
1
sin−4 (x) dx =
dx
=
tan3 (x) sec3 (x) dx = type (d).
sin4 (x) cos4 (x)
Z
Z
sin2 (x) 1
2
2
sin (x) cos(x) tan (x) csc(x) dx =
sin2 (x) cos(x)
dx
cos2 (x) sin(x)
Z
=
sin3 (x) cos(x)−1 dx = type (b).
• Rewrite using the identities: sin2 (x) = 21 (1− cos(2x)), cos2 (x) =
sin(x) cos(x) = 21 sin(2x). For example:
Z
Z
6
sin (x) dx =
( 12 (1− cos(2x))3 dx
=
=
1
8
1
8
Z
Z
1
2 (1+ cos(2x)),
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx
1 − 3 cos(2x) + 32 (1+ cos(4x)) − cos3 (2x) dx ,
where we used the binomial formula (a+b)3 = a3 + 3a2 b + 3ab2 + b3 . Now each
term can be done as type (a) with the substitution u = sin(2x) or u = sin(4x).
Recalcitrant cases. For the really tough ones, we need tricks.
example: Recall this amazing trick from §6.2:‡
Z
Z
sec(x) + tan(x)
sec(x) dx =
sec(x)
dx
sec(x) + tan(x)
Z
1
=
· sec2 (x) + sec(x) tan(x) dx
sec(x) + tan(x)
Z
=
1
du for
u
u = sec(x) + tan(x)
du = (sec2 (x) + sec(x) tan(x))dx
= ln|u| = lnsec(x)+ tan(x) .
R
R
Another trick for this is to write sec(x) dx = cos12 (x) cos(x) dx, and substitute u =
R 1
sin(x) to get 1−u
2 du. We will see how to integrate such rational functions in §7.4.
‡
The integral of secant was an important problem for map-makers in the 1600’s, when Calculus was
first developed. It calibrates the stretching in the Mercator projection, in which map directions match
compass directions.
example: Here is a tricky integration by parts, in which we get back to the same
integral we started with:
Z
Z
Z
Z
sec3 (x) dx =
sec(x)(1 + tan2 (x)) dx =
sec(x) dx + tan(x) sec(x) tan(x) dx
| {z } |
{z
}
u
Z
= ln|sec(x)+ tan(x)| + tan(x) sec(x) −
| {z } | {z }
u
Since we have
R
sec3 (x) dx
Z
v
dv
sec(x) sec2 (x) dx
| {z } | {z }
v
du
on both sides, we can solve for it to get:
sec3 (x) dx =
1
2
lnsec(x)+ tan(x) + tan(x) sec(x) .
Integrals with inside coefficients. Recall the identities:
cos(a+b) = cos(a) cos(b) − sin(a) sin(b)
cos(a−b) = cos(a) cos(b) + sin(a) sin(b)
1
2
cos(a+b) + cos(a−b)
= cos(a) cos(b).
This allows us to do integrals of the form:
Z
Z
1
cos(nx) cos(mx) dx =
cos(nx+mx)
+
cos(nx−mx)
dx
2
=
1
2(n+m)
sin (n+m)x +
1
2(n−m)
sin (n−m)x .
Some similarly useful idenitities:
1
2
1
2
cos(a−b) − cos(a+b) = sin(a) sin(b)
sin(a+b) − sin(a−b) = sin(a) cos(b).