Download Trigonometric integrals (Sect. 8.2) Product of sines and cosines

Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Substitute u = cos(x), so du = − sin(x) dx, hence
Z
I = − (1 − u 2 )k u n du.
We now need to integrate a polynomial.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Substitute u = sin(x), so du = cos(x) dx, hence
Z
I = u m (1 − u 2 )k du.
Again, we now need to integrate a polynomial.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
I = sin (x) cos (x) dx =
sin2 (x)
cos2 (x) dx.
Now use the identities
sin2 (x) =
1
1 − cos(2x) ,
2
cos2 (x) =
1
1 + cos(2x) .
2
Depending whether k or ` are odd, repeat (a), (b) or (c).
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
sin2 (x) sin(x) dx
I = sin4+1 (x) dx =
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Z
Z
2 2
1 − u ) du = − (1 − 2u 2 + u 4 ) du.
I =−
u3 u5
I = −u + 2
−
+ c.
3
5
2
1
We conclude I = − cos(x) + cos3 (x) − cos5 (x) + c.
3
5
C
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z
Z 3
3
1
2
I =
sin (x) dx =
[1 − cos(2x)] dx
2
1
I =
8
Z
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx.
Z
The first two terms are:
(1 − 3 cos(2x)) dx = x −
3
sin(2x).
2
The third term can be integrated as follows,
Z
Z
1
3
1
2
3 cos (2x) dx = 3
(1 + cos(4x)) dx =
x + sin(4x) .
2
2
4
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
2
1 − sin2 (2x) cos(2x) dx.
J = cos (2x) cos(2x) dx =
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Z
1
1
u3 1
1
2
(1 − u ) du =
J=
u−
= sin(2x) − sin3 (2x).
2
2
3
2
6
i
1h
3
3
3
1
1 3
I = x − sin(2x) + x + sin(4x) − sin(2x) + sin (2x) + c.
8
2
2
8
2
6
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Example
Z
Evaluate I =
π/8 p
1 + cos(8x) dx.
0
Solution: Use that : 1 + cos(8x) = 2 cos2 (4x). Hence,
√
√
π/8
√ Z π/8
2
2
⇒ I =
.
sin(4x)
I = 2
cos(4x) dx =
4
4
0
0
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem. These identities can be used to compute
Z
I = tan2k (x) dx,
k ∈ N.
Example
Z
Evaluate I =
tan2 (x) dx.
Solution: The identity on the far left above implies
Z
I =
tan0 (x) − 1 dx ⇒ I = tan(x) − x + c.
C
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
I = tan(2k−2) (x) tan2 (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
In the first term on the right, u = tan(x), then du = tan0 (x) dx,
Z
Z
u (2k−1)
(2k−2)
0
(2k−2)
tan
(x) tan (x) dx = u
du =
.
(2k − 1)
Z
1
(2k−1)
C
I =
tan
(x) − tan2(k−1) (x) dx.
(2k − 1)
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
I = sec(x) sec2 (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx ⇒ du = sec0 (x) dx, v = tan(x).
Z
I = sec(x) tan(x) − tan(x) sec0 (x) dx.
Recall: sec0 (x) =
sin(x)
= sec(x) tan(x).
cos2 (x)
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
tan(x) sec0 (x) dx, and we also
know sec0 (x) = sec(x) tan(x).
Z
I = sec(x) tan(x) − sec(x) tan2 (x) dx
Z
I = sec(x) tan(x) − sec(x) (sec2 (x) − 1) dx
Z
Z
sec3 (x) dx = sec(x) tan(x) +
sec3 (x) dx =
Z
Z
sec(x) dx −
sec3 (x) dx.
1
1
sec(x) tan(x) + ln sec(x) + tan(x) + c.
2
2
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Proof:
Z
Z
I =
sec(x) dx =
1
dx
cos(x)
Z
[1 + sin(x)] cos(x)
1
dx
cos(x)
cos(x) [1 + sin(x)]
Z
[1 + sin(x)]
1
I =
dx
[1 + sin(x)] cos2 (x)
cos(x)
Z 1
[1 + sin(x)] 0
I =
[1 + sin(x)] dx
cos(x)
I =
cos(x)
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Z [1 + sin(x)] 0
1
I =
[1 + sin(x)] dx
cos(x)
cos(x)
Z
0
1
dx
I =
sec(x) + tan(x)
sec(x) + tan(x)
Substitute u = sec(x) + tan(x), then
Z
du
= ln(u) + c.
I =
u
So we obtain the formula,
Z
sec(x) dx = ln sec(x) + tan(x) + c.
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Products of sines and cosines
Remark: The identities
1
cos(θ − φ) − cos(θ + φ)
2
1
sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ)
2
1
cos(θ) cos(φ) = cos(θ − φ) + cos(θ + φ) .
2
sin(θ) sin(φ) =
can be used to compute integrals of the form
Z
Z
sin(mx) sin(nx) dx,
sin(mx) cos(nx) dx,
Z
cos(mx) cos(nx) dx.
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
1
Solution: Recall: sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ) .
2
The formula above implies,
Z
1 I =
sin((3 − 4)x) + sin((3 + 4)x) dx,
2
that is,
1
I =
2
Z
− sin(x) + sin(7x) dx.
This integral is simple to do,
i
1h
1
I = cos(x) − cos(7x) + c.
2
7
C