MATH-101 Calculus and Analytical Geometry

Trigonometric Integration

Dr. Yasir Ali (yali@ceme.nust.edu.pk)

DBS&H, CEME-NUST

December 23, 2020

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 1 / 28
 Trigonometric Integration

Trigonometric Integration

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 2 / 28
 Trigonometric Integration

Trigonometric Substitutions

√ For Substitute
2 2
√a − x x = a sin θ
2 2
√a + x x = a tan θ
x 2 − a2 x = a sec θ

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 3 / 28
 Trigonometric Integration

Z
dx
I= √
a2 + x2

Put x = a tan θ =⇒ dx = a sec2 θdθ

Z
dx
I= √
a2 + x2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 4 / 28
 Trigonometric Integration

Z
dx
I= √
a2 + x2

Put x = a tan θ =⇒ dx = a sec2 θdθ

a sec2 θdθ
Z Z
dx
I= √ =⇒ I = √
a2 + x2 a2 + a2 tan2 θ

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 4 / 28
 Trigonometric Integration

Z
dx
I= √
a2 + x2

Put x = a tan θ =⇒ dx = a sec2 θdθ

a sec2 θdθ
Z Z
dx
I= √ =⇒ I = √
a2 + x2 a2 + a2 tan2 θ

p a sec2 θ
a2 + a2 tan2 θ = a sec θ =⇒ √ = sec θ
a2 + a2 tan2 θ

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 4 / 28
 Trigonometric Integration

Z
dx
I= √
a2 + x2

Put x = a tan θ =⇒ dx = a sec2 θdθ

a sec2 θdθ
Z Z
dx
I= √ =⇒ I = √
a2 + x2 a2 + a2 tan2 θ

p a sec2 θ
a2 + a2 tan2 θ = a sec θ =⇒ √ = sec θ
a2 + a2 tan2 θ

Thus Z
I= sec θdθ =⇒ ln | sec θ + tan θ|

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 4 / 28
 Trigonometric Integration

Z p
I= a2 − x2 dx

Put x = a sin θ =⇒ dx = a cos θ

Z p Z q
a2 − a2 sin2 θa cos θdθ = a2 1 − sin2 θ a cos θdθ


I=

a2
Z Z Z
2 2
= a cos θ · a cos θdθ = a cos θdθ = (1 + cos 2θ) dθ
2
a2 a2 a2 
 
sin 2θ p 
I= θ+ = (θ + sin θ cos θ) = θ + sin θ 1 − sin2 θ
2 2 2 2
r !
x a2 −1 x x  x 2
Using sin θ = a to I = sin + 1−
2 a a a

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 5 / 28
 Trigonometric Integration

Z 1p
I= 1 + x2 dx
0

π
Put x = tan θ gives x = 0 =⇒ θ = 0 and x = 1 =⇒ θ = .
4
dx = sec2 θdθ
Z θ= π4 p Z π
4
2
I= 2
1 + tan θ sec θdθ =⇒ I = sec3 θdθ
θ=0 0

π
1 4 1 √ √ 
I= (sec θ tan θ + ln | sec θ + tan θ|) = 2 + ln( 2 + 1) ' 1.148
2 0 2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 6 / 28
 Reduction Formulas

Reduction Formulas

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 7 / 28
 Reduction Formulas

n−1
Z Z
n 1 n−1
sin xdx = − sin x cos x + sinn−2 xdx
n n
n−1
Z Z
n 1 n−1
cos xdx = cos x sin x + cosn−2 xdx
n n

xn eax n
Z Z
n ax
x e dx = − xn−1 eax dx
a a
xm+1 (ln x)n
Z Z
n n
m
x (ln x) dx = − xm+1 (ln x)n−1 dx
m+1 m+1

xn cos ax n n−1 n(n − 1)

Z Z
xn sin axdx = − + 2x sin ax − xn−2 sin axdx
a a a2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 8 / 28
 Reduction Formulas

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 9 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1
sin xdx = − sin x cos x + dx =
2 2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2
sin3 xdx = − sin2 x cos x + cos x
3 3

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2 1
sin3 xdx = − sin2 x cos x + cos x = cos3 x − cos x
3 3 3

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2 1
sin3 xdx = − sin2 x cos x + cos x = cos3 x − cos x
3 3 3
Z
sin4 xdx

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2 1
sin3 xdx = −
sin2 x cos x + cos x = cos3 x − cos x
3 3 3
Z Z
4 1 3 3 2
sin xdx = − sin x cos x + sin xdx
4 4

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2 1
sin3 xdx = −
sin2 x cos x + cos x = cos3 x − cos x
3 3 3
Z Z
4 1 3 3 2
sin xdx = − sin x cos x + sin xdx
4 4
 
1 3 1 1
= − sin3 x cos x + − sin x cos x + x
4 4 2 2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 10 / 28
 Reduction Formulas

n−1
Z Z
1
sinn xdx = − sinn−1 x cos x + sinn−2 xdx
n n

Z Z
2 1 1 1 1
sin xdx = − sin x cos x + dx = − sin x cos x + x
2 2 2 2
Z
1 2 1
sin3 xdx = −
sin2 x cos x + cos x = cos3 x − cos x
3 3 3
Z Z
4 1 3 3 2
sin xdx = − sin x cos x + sin xdx
4 4
 
1 3 1 1
= − sin3 x cos x + − sin x cos x + x
4 4 2 2
3 3 1
= x− sin 2x − sin3 x cos x
8 16 4
Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)
Cal&AG December 23, 2020 10 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Powers of Sines and Cosines, Trick Using Pascal’s Triangle

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 11 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Even Cosine Powers

Figure: Pascal’s Triangle

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 12 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Even Cosine Powers

• Write 20, 30, 12, 1

20 is in the middle of
Pascal’s triangle. The
add 15s on left and right
of 20. Then add two 6s
Figure: Pascal’s Triangle and the two 1s

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 12 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Even Cosine Powers

• Write 20, 30, 12, 1

20 is in the middle of
Pascal’s triangle. The
add 15s on left and right
of 20. Then add two 6s
Figure: Pascal’s Triangle and the two 1s

(2 cos x)6 = 20 cos 0 + 30 cos 2x + 12 cos 4x + 2 cos 6x 26 = 64

1
cos6 x = (20 cos 0x + 30 cos 2x + 12 cos 4x + 2 cos 6x)
64

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 12 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

1
cos6 x = (10 + 15 cos 2x + 6 cos 4x + cos 6x)
32

Z Z
6 1
I = cos x = (10 + 15 cos 2x + 6 cos 4x + cos 6x) dx
32
 
1 15 6 1
= 10x + sin 2x + sin 4x + sin 6x + C
32 2 4 6

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 13 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Odd Cosine Powers

Write 20, 10, 2

(2 cos x)5 = 20 cos x+10 cos 3x+2 cos 5x

1
cos5 x = (20 cos x + 10 cos 3x + 2 cos 5x)
32

Z Z
1
I= cos5 x = (20 cos x + 10 cos 3x + 2 cos 5x) dx
32
 
1 5 1
= 10 sin x + sin 3x + sin 5x + C
16 3 5

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 14 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine
functions,

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine
functions,
3 Even powers of sin x
get expanded in
terms of cosine
functions.

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine 1
functions, sin2 x = (1 − cos 2x)
2
3 Even powers of sin x
get expanded in
terms of cosine
functions.

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine 1
functions, sin2 x = (1 − cos 2x)
2
3 Even powers of sin x 3 1
sin3 x = sin x − sin 3x
get expanded in 4 4
terms of cosine
functions.

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine 1
functions, sin2 x = (1 − cos 2x)
2
3 Even powers of sin x 3 1
sin3 x = sin x − sin 3x
get expanded in 4 4
3 1 1
terms of cosine sin4 x = − cos 2x + cos 4x
8 2 2
functions.

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Trick Using Pascal’s Triangle for Sine Powers

1 Alternative sign
starting from +
2 Odd powers of sin x
get expanded in
terms of sine 1
functions, sin2 x = (1 − cos 2x)
2
3 Even powers of sin x 3 1
sin3 x = sin x − sin 3x
get expanded in 4 4
3 1 1
terms of cosine sin4 x = − cos 2x + cos 4x
8 2 2
functions. 5 5 1
sin5 x = sin x − sin 3x + sin 5x
8 16 16

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 15 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 16 / 28
 Reduction Formulas Trick Using Pascal’s Triangle

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 17 / 28
 Products of Powers of Sines and Cosines

Products of Powers of Sines and Cosines

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 18 / 28
 Products of Powers of Sines and Cosines

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 19 / 28
 Products of Powers of Sines and Cosines

Z
I= sin4 x cos5 xdx, n= 5 is odd

So
Z Z
4 4

2
I= sin x cos x cos xdx =⇒ I = sin4 x 1 − sin2 x cos xdx

Put u = sin x impliesdu = cos xdx, thus

Z Z
4 2 2
I = u (1 − u ) du =⇒ I = (u4 − 2u6 + u8 )du

u5 2u7 u9 sin5 x 2 sin7 x sin9 x

I= − + +C = I = − + +C
5 7 9 5 7 9
Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)
Cal&AG December 23, 2020 20 / 28
 Products of Powers of Sines and Cosines

Z Z
4 4

2
2
I= sin x cos xdx = sin2 x cos2 x

1−cos 2x 1+cos 2x
Use sin2 x = 2 and cos2 x = 2

Z  2  2
1 − cos 2x
Z
1 + cos 2x 1
I= dx = (1 − cos2 2x)2 dx
2 2 16

Z Z
1 4 1
I= sin 2xdx letting 2x = u =⇒ I = sin4 udu
16 32
Use reduction formula to get the answer.

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 21 / 28
 Products of Powers of Sines and Cosines

Integrate sin3 x cos2 x

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 22 / 28
 Products of Powers of Sines and Cosines

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 23 / 28
 Products of Powers of Sines and Cosines

Z
I= tan2 x sec4 xdx

As power of sec is even, that is, n = 2 so Split off a factor of sec2 x.

Put u = tan x =⇒ du = sec2 xdx .
Z Z
I = tan x sec x sec xdx = tan2 x tan2 x + 1 sec2 xdx
2 2

2

Z
1 1
I= u2 (u2 + 1)du = u5 + u3 + C
5 3

1 1
I= tan5 x + tan3 x + C
5 3
Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)
Cal&AG December 23, 2020 24 / 28
 Products of Powers of Sines and Cosines

Z
tan3 x sec3 xdx

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 25 / 28
 Products of Powers of Sines and Cosines

Z
tan2 x sec xdx

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 26 / 28
 Products of Powers of Sines and Cosines

1
sin α cos β = [sin(α − β) + sin(α + β)]
2
1
sin α sin β = [cos(α − β) − cos(α + β)]
2
1
cos α cos β = [cos(α − β) + cos(α + β)]
2

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 27 / 28
 Products of Powers of Sines and Cosines

1
sin α cos β = [sin(α − β) + sin(α + β)]
2
1
sin α sin β = [cos(α − β) − cos(α + β)]
2
1
cos α cos β = [cos(α − β) + cos(α + β)]
2

Z Z
1
sin 7x cos 3xdx = [sin(7x − 3x) + sin(7x + 3x)] dx
2
Z
1 1
= (sin 4x + sin 10x)dx = − cos 4x − cos 10x + C
8 20

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 27 / 28
 Products of Powers of Sines and Cosines

Summary

Trigonometric Substitutions
Powers of sine and Cosine Functions
Product of Powers of x and Transcendental Functions
Product of Powers of Sine and Cosine
Product of Powers of Tan and Secant

Dr. Yasir Ali (yali@ceme.nust.edu.pk) (DBS&H, CEME-NUST)

Cal&AG December 23, 2020 28 / 28