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A lecture on trigonometric integrals and
trigonometric substitution∗
alozano†
2013-03-21 19:58:03
1
Trigonometric Integrals
First, we must recall a few trigonometric identities:
sin2 x + cos2 x
2
sec x
=
1
(1)
2
1 + tan x
1 − cos(2x)
sin2 x =
2
1
+
cos(2x)
cos2 x =
2
sin(2x) = 2 sin x cos x
cos(2x)
=
=
2
2
cos x − sin x.
(2)
(3)
(4)
(5)
(6)
The most usual integrals which involve trigonometric functions can be solved
using the identities above.
R
R
Example 1.1. sin xdx = − cos x+C and cos xdx = sin x+C are immediate
integrals.
R
R
Example 1.2. For sin2 xdx, cos2 xdx we use formulas (3) and (4) respectively, e.g.
Z
Z
Z
1 − cos(2x)
1
1
sin(2x)
sin2 xdx =
dx =
(1 − cos(2x))dx =
x−
+ C.
2
2
2
2
R
R
Example 1.3. For integrals of the form cosm x sin x dx or sinm x cos x dx
we use substitution with u = cos x or u = sin x respectively, e.g.
Z
Z
u3
cos3 x
cos2 x sin xdx = −u2 du = − +C = −
+C. [u = cos x, du = − sin xdx]
3
3
∗ hALectureOnTrigonometricIntegralsAndTrigonometricSubstitutioni
created: h2013-0321i by: halozanoi version: h37576i Privacy setting: h1i hFeaturei h26A36i
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compatible with the CC-BY-SA license.
1
In the following examples, we use equations (1) in the forms sin2 x = 1 −
cos x or cos2 x = 1 − sin2 x to transform the integral into one of the type
described in Example ??.
2
Example 1.4.
Z
sin3 xdx
Z
Z
sin2 x sin xdx = (1 − cos2 x) sin xdx
Z
Z
=
sin xdx − cos2 x sin xdx
=
= − cos x +
Similarly one can solve
Example 1.5.
Z
cos3 x sin2 xdx
R
cos3 x
+ C.
3
cos3 xdx.
Z
2
Z
cos x cos x sin xdx = (1 − sin2 x) cos x sin2 xdx
Z
Z
=
cos x sin2 xdx − cos x sin4 xdx
2
=
sin3 x sin5 x
−
+ C.
3
5
R
R
1.6. In order to solve cos5 x sin3 xdx we express it first as cos5 x sin2 x sin x =
RExample
cos5 x(1 − cos2 x) sin xdx and then proceed as in the previous example.
=
One can use similar tricks to solve integrals which involve products of powers
of sec x and tan x, by using Equation (2). Also, recall that the derivative of tan x
is sec2 x while the derivative of sec x is sec x tan x.
Example 1.7.
Z
tan5 x sec4 xdx
Z
2
2
Z
tan x sec x sec xdx = tan5 x(1 + tan2 x) sec2 xdx
Z
Z
=
tan5 x sec2 xdx + tan7 x sec2 xdx
=
=
Example 1.8.
Z
tan3 x sec4 xdx
5
tan6 x tan8 x
+
+ C.
6
8
Z
=
4
Z
tan x tan x sec xdx = tan x(sec2 x − 1) sec4 xdx
Z
Z
=
tan x sec x sec5 xdx − tan x sec x sec3 xdx
=
2
sec6 x sec4 x
−
+ C.
6
4
2
2
Trigonometric Substitutions
R1√
One can easily deduce that 0 1 − x2 dx has value π4 . Why? Simply because
√
the graph of the function y = 1 − x2 is√half a circumference of radius r = 1
(because if you square both sides of y = 1 − x2 you obtain x2 + y 2 = 1 which
is the equation of a circle or radius r = 1). Therefore, the area under the graph
is a quarter of the area of a circle.
R1√
How does one compute 0 1 − x2 dx without using the geometry of the
problem? This is the prototype of integral where a trigonometric substitution
will work very nicely. Notice that neither substitution nor integration by parts
will work appropriately.
R1√
Example 2.1. Suppose we want to solve 0 1 − x2 dx with analytic methods.
We will use a substitution x = sin θ (so θ will be our new
pvariable of integration),
√
because, as we know from Equation (1), 1 − x2 = 1 − sin2 θ = cos θ, thus
getting rid of the pesky square root. Notice that dx = cos θdθ. We also need to
find the new limits of integration with respect to the new variable of integration,
namely θ. When x = 0 = sin θ we must have θ = 0. Similarly, when x = 1 =
sin θ one has θ = π/2. We are now ready to integrate:
Z
1
p
1 − x2 dx
π/2
Z
=
π/2
Z
cos2 θdθ
(cos θ) cos θdθ =
0
0
0
π/2
Z
=
0
1
1 + cos(2θ)
dθ =
2
2
π/2
sin(2θ)
= π/4.
θ+
2
0
Notice that we made use of Equation (4) in the second line.
Rr√
Example 2.2. Similarly, one can solve 0 r2 − x2 dx by using a substitution
p
√
x = r sin θ. Indeed, r2 − x2 = r2 − r2 sin2 θ = r cos θ and dx = r cos θdθ.
The limits of integration with respect to θ are again θ = 0 to θ = π/2 (check
this!). Thus:
Z
r
p
r2
−
x2 dx
Z
π/2
2
=
0
r (cos θ) cos θdθ = r
2
Z
0
=
r2
π/2
cos2 θdθ
0
Z
0
π/2
r2
1 + cos(2θ)
dθ =
2
2
θ+
sin(2θ)
2
π/2
= r2 π/4.
0
Thus, we have proved that a quarter of a circle of radius r has area r2 π/4 which
implies that the area of such a circle is πr2 , as usual.
√
substitutions usually work when expressions like r2 − x2 ,
√ The trigonometric
√
r2 + x2 , x2 − r2 appear in the integral at hand, for some real number r. Here
is a table of the suggested change of variables in each particular case:
3
If you see...
√
1 − x2
√
2
2
√r − x
2
√1+x
2
2
√r + x
2
√x −1
x2 − r 2
try this...
x = sin θ
x = r sin θ
x = tan θ
x = r tan θ
x = sec θ
x = r sin θ
p because...
2
p 1 − sin θ = cos θ
2
2
√r − sin θ = r cos θ
2
√ 1 + tan θ = sec θ
2
2
r + tan θ = r sec θ
√
2
√ sec θ − 1 = tan θ
2
sec θ − 1 = r tan θ
Remark 2.3. The above are “suggested” substitutions,
R √ they may not be the
most ideal choice! For example, for the integral 2x 1 − x2 dx, the change
u = 1 − x2 will work much better than x = sin θ.
Example 2.4. We would like to find the value of
Z 2
1
√
dx.
√
3
x2 − 1
2 x
Since neither a u-substitution nor integration
√ by parts seem appropriate, we try
x = sec θ, dx = sec θ tan θdθ. When x = 2 = sec θ one has θ = π/4 while
x = 2 implies θ = π/3. Hence:
Z
2
√
2
x3
√
1
dx
x2 − 1
Z
π/3
=
π/4
sec θ tan θ
dθ =
sec3 θ tan θ
Z
π/3
π/4
1
dθ =
sec2 θ
and the last integral is easy to compute using Equation (4).
4
Z
π/3
π/4
cos2 θdθ