Download Differential and Integral Calculus

Integration Techniques
Differential and Integral Calculus
Prof. Lahcen Laayouni
School of science and engineering
El Akhawayn University
Trigonometric Integrals
Friday, April 25th , 2008
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Outline
1
Integration Techniques
Trigonometric Integrals
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Example 1:
Evaluate the indefinite integral
Z
sin2 3xdx .
Solution:
Using the half-angle identity, we get
Z
Z 1 − cos 6x
sin2 3xdx =
dx
2
1
1
=
x − sin 6x + C
2
6
Example 2:
Evaluate the integral
Z
sin3 x cos x dx .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Solution:
Setting u = sin x , so du = cos x dx , thus
Z
Z
1
sin3 x cos x dx =
u 3 du = u 4 + C
4
1
=
sin4 x + C.
4
Example 3:
Evaluate
Z
sin5 x cos2 x dx .
Solution:
We separate sine factor and rewrite the remaining sin4 x factor in terms
of cos x , we obtain
sin5 x cos2 x = (sin2 x)2 cos2 x sin x = (1 − cos2 x)2 cos2 x sin x .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Solution:(Continue)
Substituting u = cos x , we have du = − sin xdx , then
Z
5
2
sin x cos x dx
=
Z
(sin2 x)2 cos2 xsin x dx
=
Z
(1 − cos2 x)2 cos2 x sin x dx
=
Z
= −
2 2 2
(1 − u ) u (−du) = −
u3 2 5 1 7
− u + u
3
5
7
Z
(u 2 − 2u 4 + u 6 )du
+C
1
2
1
= − cos3 x + cos5 x − cos7 x + C.
3
5
7
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Strategy for evaluating
Z
sinm x cosn x dx :
If the power of cosine is odd (n = 2k + 1) , save one cosine factor
and use cos2 x = 1 − sin2 x to express the remaining factors in
terms of sine
Z
Z
m
n
sinm x cos2k +1 x dx
sin x cos x dx =
=
Z
sinm x(cos2 )k cos x dx
=
Z
sinm (1 − sin2 x)k cos x dx
Then substitute u = sin x .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Strategy for evaluating
Z
sinm x cosn x dx :
If the power of sine is odd (m = 2k + 1) , save one sine factor and
use sin2 x = 1 − cos2 x to express the remaining factors in terms
of cosine
Z
Z
m
n
sin2k +1 x cosn x dx
sin x cos x dx =
=
Z
(sin2 )k x cosn sin x dx
=
Z
(1 − cos2 x)k cosn x sin x dx
Then substitute u = cos x .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Strategy for evaluating
Z
sinm x cosn x dx :
If the powers of both sine and cosine are even, use the half-angle
identities
sin2 x =
1
(1 − cos 2x) ,
2
cos2 x =
1
(1 + cos 2x)
2
It is sometimes helpful to use the identity
sin x cos x =
1
sin 2x .
2
Example 4:
Evaluate
Z
tan6 x sec4 x dx .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Solution:
Separating one sec2 x factor and using the identity
sec2 x = 1 + tan2 x , we obtain
Z
6
4
tan x sec x dx
=
Z
tan6 x sec2 x sec2 x dx
=
Z
tan6 x(1 + tan2 x) sec2 x dx
Substituting u = tan x , so du = sec2 xdx , then
Z
Z
Z
6
4
6
2
tan x sec x dx =
u (1 + u )du = (u 6 + u 8 )du
=
u7 u9
1
1
+
+ C = tan7 x + tan9 x + C
7
9
7
9
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Example 5:
Evaluate
Z
tan5 x sec7 x dx .
Solution:
Rewriting the integral and using the identity tan2 x = sec2 x − 1 , we
obtain
Z
Z
5
7
tan4 x sec6 x sec x tan x dx
tan x sec x dx =
=
Z
(sec2 x − 1)2 sec6 x sec x tan x dx.
Setting u = sec x , so du = sec x tan x dx , then
Z
Z
5
7
tan x sec x dx =
(u 2 − 1)2 u 6 du.
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Solution:
Thus
Z
Z
5
7
tan x sec x dx =
(u 4 − 2u 2 + 1)u 6 du
=
Z
=
u 11
u9 u7
−2 +
+C
11
9
7
=
1
2
1
sec11 x − sec9 x + sec7 x + C.
11
9
7
(u 10 − 2u 8 + u 6 )du
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Strategy for evaluating
Z
tanm x secn x dx
If the power of secant is even (n = 2k, k ≥ 2) , save a factor of
sec2 x and use sec2 x = 1 + tan2 x to express the remaining
factors in terms of tan x
Z
Z
m
n
tan x sec x dx =
tanm x sec2k x dx
=
Z
tanm x(sec2 x)k −1 sec2 x dx
=
Z
tanm x(1 + tan2 x)k −1 sec2 x dx
Then substitute u = tan x .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Strategy for evaluating
Z
tanm x secn x dx
If the power of tangent is odd (m = 2k + 1) , save a factor of
sec x tan x and use tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x
Z
Z
tanm x secn x dx =
tan2k +1 x secn x dx
=
Z
(tan2 x)k secn−1 sec x tan x dx
=
Z
(sec2 x − 1)k secn−1 x sec x tan x dx
Then substitute u = sec x .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Example 6:
Evaluate the indefinite integral
Z
sec x dx .
Solution:
First, we rewrite the secant term as follows
sec x =
1
cos x
cos x
=
.
=
2
cos x
cos x
1 − sin2 x
Next we use the algebraic identity
1
1
1
1
=
+
.
2 1+x
1−x
1 − x2
then
2 cos x
2
1 − sin x
=
cos x
cos x
+
.
1 − sin x
1 + sin x
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Solution:
Therefore Z
sec x dx =
=
1
2
Z cos x
cos x
+
1 + sin x
1 − sin x
dx
1
[ln |1 + sin x| − ln |1 − sin x|] + C
2
or
Z
sec x dx
1 1 + sin x 1 (1 + sin x)2 ln +
C
=
ln
+C
2
1 − sin x 2 1 − sin2 x (1 + sin x)2 1/2
1 + sin x +C
= ln + C = ln cos x cos2 x =
= ln |sec x + tan x| + C.
Z
Similarly, we can show that
csc x dx = − ln | csc x + cot x| + C .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Example 7:
Evaluate the integral
Z
sin 3x cos 2x dx .
Solution:
Applying the trigonometry identity
sin A cos B =
we get
Z
sin 3x cos 2x dx
=
=
=
1
[sin(A − B) + sin(A + B)] ,
2
Z
1
[sin(3 − 2)x + sin(2 + 3)x] dx
2
Z
1
[sin x + sin(5x)] dx
2
1
1
− cos x − cos 5x + C.
2
5
Prof. Lahcen Laayouni
Differential and Integral Calculus
Integration Techniques
Trigonometric Integrals
Integration Techniques
Trigonometric identities:
For all A and B , we have
sin A sin B
=
sin A cos B
=
cos A cos B =
1
[cos(A − B) − cos(A + B)]
2
1
[sin(A − B) + sin(A + B)]
2
1
[cos(A − B) + cos(A + B)]
2
Prof. Lahcen Laayouni
Differential and Integral Calculus