Download Trigonometric Integrals

Calculus 2
Lia Vas
Trigonometric Integrals
Let us consider the integrals of the form
Z
f (sin x) cos xdx
Z
or
f (cos x) sin xdx
where f (x) is a function with antiderivative F (x). Using the substitution
u = sin x for the first integral
and
u = cos x for the second integral
R
one can reduce the first integral to f (u)du = F (u) + c = F (sin x) + c and the second integral to
f (u)(−du) = −F (u) + c = −F (cos x) + c.
R
This idea can be applied to the integrals of the form
Z
sinn x cosm x dx
if either m or n are odd. We will refer to this situation as the good case.
If n is odd rewrite sinn−1 x as a function of cos x using
the trigonometric identity sin2 x =
R
2
1−cos x. Note that then you obtain an integral of the form f (cos x) sin xdx where f is a polynomial
function (thus easy to integrate). You can evaluate the integral using the substitution
u = cos x.
If m is odd rewrite cosm−1 x as a function of sin x using
the trigonometric identity cos2 x =
R
2
1−sin x. Note that then you obtain an integral of the form f (sin x) cos xdx where f is a polynomial
function. You can evaluate the integral using the substitution
u = sin x.
Let us consider the case when both n or m are even. Let us refer to this as the bad case (although
it is not that bad). The idea is to use one or more of the following three trigonometric identities
1
sin2 x = (1 − cos 2x)
2
1
cos2 x = (1 + cos 2x)
2
and
sin x cos x =
1
sin 2x
2
to reduce the integral to a sum of integrals in which the powers of cosines and sines are at most 1.
Then you can integrate term by term.
If you encounter a multiple of x in the argument of sin or cos, note that the three identities above
become sin2 ax = 21 (1 − cos 2ax), cos2 ax = 12 (1 + cos 2ax), and sin ax cos ax = 12 sin 2ax.
Note: If you have to integrate other trigonometric functions, you can convert them to sin and cos
functions using the trigonometric identities.
1
Practice Problems. Evaluate the following integrals:
1.
Z
5.
9.
Z
Z
10
sin x cos x dx
tan x dx
sin5 x dx
2.
6.
10.
Z
Z
Z
3
2
sin x cos x dx
3.
2
cos x tan x dx
7.
Z
Z
e
cos x
sin x dx
2
sin x dx
4.
8.
Z
Z
cos x
dx
1 + sin2 x
sin2 x cos2 x dx
cos4 x dx
11. Finding the center of mass. Let R be the region
between the graphs of f and g such that
Rb
f (x) ≥ g(x) on interval [a, b]. The area A of R is A = a (f (x) − g(x)) dx. Then the center of mass
of R is the point (x̄, ȳ) where
x̄ =
1 Zb
x (f (x) − g(x)) dx
A a
1 Zb1
((f (x))2 − (g(x))2 ) dx
ȳ =
A a 2
Find the center of mass of the region bounded by the given curves.
(a) y = sin x, y = cos x, x = 0, x = π/4.
(b) y = sin 2x, y = 0, x = 0, x = π/2.
Solutions.
R
1. This integral is already
in the form
f (sin x) cos xdx so use the substitution u = sin x. The
R
R 10
du
1 11
1
integral becomes u10 cos x cos
=
u
du = 11
u + c = 11
sin11 x + c.
x
2. The sine function has the odd
power. So, this is Rthe “good case”. Write sin3 x as sinR2 x sin x =
R
(1 − cos2 x) sin x and get cos2 x sin3 x dx =R cos2 x (1 − cos2 x) sin x dx = (cos2 x −
cos4 x) sin x dx. This integral is of the form f (cos x) sin x dx that
can be evaluated using
R
2
u = cos x ⇒ du = − sin xdx ⇒ − du
=
dx.
The
integral
reduces
to
(cos
x−cos4 x) sin x dx =
sin x
R 2
R
(u − u4 ) sin x − du
= − (u2 − u4 )du = −1
u3 + 51 u5 + c = −1
cos3 x + 15 cos5 x + c.
sin x
3
3
R
3. This integral
is of the form R f (cos x) sin xdx so use the substitution u = cos x. The integral
R u
becomes e sin x − du
= − eu du = −eu + c = −ecos x + c.
sin x
R
4. This integral
is of the form f (sin x) cos xdx so use the substitution u = sin x. The integral
R cos du
1
−1
becomes 1+u2 cos x = 1+u
u + c = tan−1 (sin x) + c.
2 du = tan
5. Write tan x
as sin x = 1 sin Rx and treat this as f (cos x) sin x. So, use the substitution u = cos x
R 1 cos x du cos x
to obtain u sin x − sin x = − u1 du = − ln |u| + c = − ln | cos x| + c.
2
6.
sin x
cos2 x tan x dx = cos2 x cos
dx = cos x sin x. Note that this is a good case with both
x
exponents equal (super good case!) so both substitutions u = cos x and u = sin x could work.
R
R
With u = sin x, one obtains
R
R
du
cos xu cos
= 12 u2 + c = 12 sin2 x + c.
x
= −1
u2 +c = −1
cos2 x+c. Note that this is the same
With u = cos x, one obtains u sin x − du
sin x
2
2
as the previous answer since −1
cos2 x + c = −1
(1 − sin2 x) + c = −1
+ 12 sin2 x + c = 21 sin2 x + c1 .
2
2
2
R
7. This is the bad case. Use the trigonometric identity sin2 x = 12 (1 − cos 2x) to have
cos 2x)dx. Use the substitution u = 2x and obtain 12 x − 14 sin(2x) + c.
R 1
(1
2
−
8. This is the
bad case as well.R Use both identities sin2 x = 21 (1−cos
2x) and cos2 x = 12 (1+cos 2x)
R
R
to have sin2 x cos2 xdx = 41 (1 − cos 2x)(1 + cos 2x)dx = 14 (1 − cos2 2x)dx. Then use the trig
with 2x instead of x Rto reduce cos2 2x to linear terms. Obtain
identity
cos2 x = 12 (1 + cos 2x)
R 1
R
1
(1 − 21 (1 + cos 4x))dx = ( 14 − 18 − 81 cos 4x)dx = ( 18 − 81 cos 4x)dx = 18 x − 32
sin 4x + c.
4
1
sin 2x first and
then sin2 2x 12 (1 − cos 4x) after. In this
Alternatively, you Rcan use sin x cos x =
2
R
R
2
1
1
1
case, you will get (sin x cos x)2 dx = 4 sin 2xdx = 8 (1 − cos 4x)dx = 18 x − 32
sin 4x + c.
9. This is the good case. sin5 x = sin4 x sin x = (sin2 x)2 sin x = (1 − cos2 x)2 sin x. Since this
is of the form f (cos x)Rsin x, you can use the substitution u = cos x. The integral becomes
R
(1 − u2 )2 sin x − du
= −(1 − 2u2 + u4 )du = −u + 23 u3 − 51 u5 = − cos x + 23 cos3 x − 51 cos5 x + c.
sin x
10. This is the bad case. Using the trig identities cos4 x = (cos2 x)2 = ( 21 (1 + cos 2x))2 = 14 (1 +
2 cos 2x + cos2 2x) = 14 (1 + 2 cos 2x + 12 (1 + cos 4x)) = 41 + 12 cos 2x + 81 + 81 cos 4x = 83 + 12 cos 2x +
1
1
cos 4x. Integrating term by term now, you obtain 83 x + 41 sin 2x + 32
sin 4x + c.
8
11. (a) Note that on (0, π4 ) cos x is larger than sin x. So the area A can be evaluated as A =
√
R π/4
π/4
= 2 − 1 = .414. The x-coordinate x̄ is x̄ =
0 (cos x − sin x)dx = (sin x + cos x)|0
1 R π/4
x(cos x−sin x) hdx. Use the integration by parts with u = x and
dv = (cos x−sin x)dx for
.414 0
i
√
π/4
π/4
1
1
( π4 2 − 1) = .267.
this integral. x̄ = .414 x(sin x + cos x)|0 − (− cos x + sin x)|0 dx = .414
R π/4
1
1
ȳ = .414
(cos2 x − sin2 x)dx. This is the “bad case”. Using the trigonometric identities
0
2
1 1 R π/4
1 R π/4
for sin2 x and cos2 x, we obtain .414
(1 + cos 2x − 1 + cos 2x)dx = 1.656
2 cos 2xdx =
0
4 0
π/4
1
sin 2x|0 = .6035. So, the center of mass is (.267, .6035).
1.656
R π/2
π/2
R π/2
(b) A = 0 sin 2xdx = −1
cos 2x|0 = −1
(−1 − 1) = 1. x̄ = 0 x sin 2xdx. Using the
2
2
R π/2
π/2
−x
integration by parts obtain that x̄ = ( 2 cos 2x + 14 sin 2x)|0 = π4 . ȳ = 0 12 sin2 2xdx. Using
R π/2
π/2
the trigonometric identities for sin2 x, we obtain 41 0 (1 − cos 4x)dx = 41 (x − 41 sin 4x)|0 = π8 .
So, the center of mass is ( π4 , π8 ).
3