Filetype PDF | Posted on 15 Sep 2022 | 3 years ago
Authentication
digital resources
167x Filetype PDF File size 0.29 MB Source: sites.math.northwestern.edu
1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 26
1.6. Trigonometric Integrals and Trigonometric
Substitutions
1.6.1. TrigonometricIntegrals. Herewediscussintegralsofpow-
ers of trigonometric functions. To that end the following half-angle
identities will be useful:
sin2 x = 1(1 −cos2x),
2
cos2x = 1(1+cos2x).
2
Remember also the identities:
2 2
sin x+cos x = 1,
2 2
sec x = 1+tan x.
1.6.1.1. Integrals of Products of Sines and Cosines. We will study
now integrals of the form
Z m n
sin xcos xdx,
including cases in which m = 0 or n = 0, i.e.:
Z cosnxdx; Z sinmxdx.
The simplest case is when either n = 1 or m = 1, in which case the
substitution u = sinx or u = cosx respectively will work.
Example: Z sin4xcosxdx = ···
(u = sinx, du = cosxdx)
Z 5 5
· · · = u4du = u +C = sin x +C .
5 5
More generally if at least one exponent is odd then we can use the
2 2
identity sin x+cos x = 1totransformtheintegrandintoanexpression
containing only one sine or one cosine.
1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 27
Example:
Z 2 xcos3xdx = Z sin2xcos2xcosxdx
sin
=Z sin2x(1−sin2x)cosxdx=···
(u = sinx, du = cosxdx)
· · · = Z u2 (1 − u2)du = Z (u2 − u4)du
u3 u5
= 3 − 5 +C
3 sin5
= sin x − x +C .
3 5
If all the exponents are even then we use the half-angle identities.
Example:
Z 2 2 Z 1 1
sin xcos xdx = 2(1 −cos2x)2(1+cos2x)dx
=1Z 2 2x)dx
4 Z (1 −cos
=1 (1−1(1+cos4x))dx
4 2
=1Z (1−cos4x)dx
8
=x−sin4x+C.
8 32
1.6.1.2. Integrals of Secants and Tangents. The integral of tanx
can be computed in the following way:
Z tanxdx=Z sinx dx=−Z du = −ln|u|+C =−ln|cosx|+C ,
cosx u
where u = cosx. Analogously
Z cotxdx = Z cosx dx = Z du = ln|u|+C = ln|sinx|+C ,
sinx u
where u = sinx.
1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 28
The integral of secx is a little tricky:
Z secxdx = Z secx(tanx+secx)dx = Z secxtanx+sec2xdx=
secx+tanx secx+tanx
Z du =ln|u|+C = ln|secx+tanx|+C ,
u
2
where u = secx+tanx, du = (secxtanx+sec x)dx.
Analogously:
Z cscxdx = −ln|cscx+cotx|+C .
More generally an integral of the form
Z m n
tan xsec xdx
can be computed in the following way:
(1) If m is odd, use u = secx, du = secxtanxdx.
2
(2) If n is even, use u = tanx, du = sec xdx.
Z 3 2
Example: tan xsec xdx = ···
Since in this case m is odd and n is even it does not matter which
method we use, so let’s use the first one:
(u = secx, du = secxtanxdx)
Z 2 Z 2
· · · = tan x secx tanxsecxdx = (u −1)udu
| {z }|{z}| {z }
u2−1 u du Z
= (u3−u)du
4 u2
=u − +C
4 2
= 1sec4x− 1sec2x+C .
4 2
Next let’s solve the same problem using the second method:
1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 29
(u = tanx, du = sec2xdx)
Z 3 2 Z 3 u4 +C = 1 4
tan x sec xdx = u du = 4 4 tan x+C .
| {z }| {z }
u3 du
Although this answer looks different from the one obtained using the
first method it is in fact equivalent to it because they differ in a con-
stant:
1 4 1 2 2 1 4 1 2 1
4 tan x = 4(sec x−1) = 4 sec x− 2 sec x+4 .
| {z }
previous answer
1.6.2. Trigonometric Substitutions. Here we study substitu-
tions of the form x = some trigonometric function.
Example: Find Z √1−x2dx.
Answer: We make x = sint, dx = costdt, hence
√ 2 p 2 √ 2
1−x = 1−sin t= cos t=cost,
and Z
Z √1−x2dx=Z cost costdt
2
= cos tdt
=Z 1(1+cos2t)dt (half-angle identity)
2
= t + sin2t +C
2 4
= t + 2sintcost +C (double-angle identity)
2 4
p 2
= t + sint 1−sin t +C
2 2
−1 √ 2
= sin x + x 1−x +C .
2 2
The following substitutions are useful in integrals containing the
following expressions:
no reviews yet
Please Login to review.
