1
Methods of Integration
1. Product of Sines and Cosines
P1: ∫ 2 sin 𝑢 cos 𝑣 𝑑𝑥 = ∫|sin(𝑢 + 𝑣) + sin(𝑢 − 𝑣)| 𝑑𝑥
P2: ∫ 2 cos 𝑢 cos 𝑣 𝑑𝑥 = ∫|cos(𝑢 + 𝑣) + cos(𝑢 − 𝑣)| 𝑑𝑥
P3: ∫ 2 sin 𝑢 sin 𝑣 𝑑𝑥 = ∫|cos(𝑢 − 𝑣) − cos(𝑢 + 𝑣)| 𝑑𝑥
Examples: Evaluate:
1. ∫ cos 6𝑥 cos 2𝑥 𝑑𝑥
1
= ∫ 2 cos 6𝑥 cos 2𝑥 𝑑𝑥
2
1
= 2 ∫|cos(6𝑥 + 2𝑥) + cos(6𝑥 − 2𝑥)| 𝑑𝑥
1
= 2
∫|cos 8𝑥 + cos 4𝑥| 𝑑𝑥
1 1
Let u = 8x, du = 8 dx, nf = 8 ; u = 4x, du = 4 dx, nf = 4
1 1 1
= ( sin 8𝑥 + sin 4𝑥) + 𝐶
2 8 4
𝟏 𝟏
= 𝐬𝐢𝐧 𝟖𝒙 + 𝐬𝐢𝐧 𝟒𝒙 + 𝑪
𝟏𝟔 𝟖
2. ∫ 3 sin 5𝑥 cos 4𝑥 𝑑𝑥
3
= ∫ 2 sin 5𝑥 cos 4𝑥 𝑑𝑥
2
3
= 2 ∫|sin(5𝑥 + 4𝑥) + sin(5𝑥 − 4𝑥)| 𝑑𝑥
3
= ∫|sin 9𝑥 + sin 𝑥| 𝑑𝑥
2
3 1
= (− cos 9𝑥 − cos 𝑥) + 𝐶
2 9
𝟏 𝟑
= − 𝐜𝐨𝐬 𝟗𝒙 − 𝐜𝐨𝐬 𝒙 + 𝑪
𝟔 𝟐
3. ∫ sin(1 − 2𝑥) sin(2𝑥 + 3) 𝑑𝑥 , u = 1 – 2x, v = 2x + 3
1
= ∫ 2 sin(1 − 2𝑥) sin(2𝑥 + 3)𝑑𝑥
2
1
= ∫|𝑐𝑜𝑠[(1 − 2𝑥) − (2𝑥 + 3)] − 𝑐𝑜𝑠[(1 − 2𝑥) + (2𝑥 + 3)]|
2
1
= ∫|cos(−4𝑥 − 2) − cos(4)| 𝑑𝑥
2
1 1
= {− sin(−4𝑥 − 2) − 𝑥𝑐𝑜𝑠 4} + 𝐶
2 4
𝟏 𝟏
= − 𝟖 𝐬𝐢𝐧(−𝟒𝒙 − 𝟐) − 𝟐 𝒙𝒄𝒐𝒔 𝟒 + 𝑪
MATH 104 – CALCULUS II (INTEGRAL CALCULUS RGREPIQUE
, 2
2. Powers of Sines and Cosines
∫ 𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 𝑑𝑥
Where: v = differentiable function of x
m, n = real numbers.
CASE 1: When m is a positive odd integer and n is any number, we may write:
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛𝑚−1 𝑣 𝑐𝑜𝑠 𝑛 𝑣) sin 𝑣
Since m is odd, then m – 1 is even and therefore, we may use the trigonometry identity,
𝑠𝑖𝑛2 𝑣 = 1 − 𝑐𝑜𝑠 2 𝑣
To express 𝑠𝑖𝑛𝑚−1 𝑣 in terms of the powers of cos v. Then the given integral is reduced to the
form,
∫(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 cos 𝑣) sin 𝑣 𝑑𝑥
which can now be evaluated by use of I4 by letting u = cos v.
CASE 2: When m is any number and n is a positive odd number, we may write:
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛−1 𝑣) cos 𝑣
Since m is odd, then m – 1 is even and therefore, we may use the trigonometry identity,
𝑐𝑜𝑠 2 𝑣 = 1 − 𝑠𝑖𝑛2 𝑣
To express 𝑠𝑖𝑛𝑚−1 𝑣 in terms of the powers of cos v. Then the given integral is reduced to the
form,
∫(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 sin 𝑣) cos 𝑣 𝑑𝑥
which can now be evaluated by use of I4 by letting u = sin v.
CASE 3: When m and n are both even integers (either both positive or one positive and one zero),
we may write:
𝑚 𝑛
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛2 𝑣 ) 2 (𝑐𝑜𝑠 2 𝑣 ) 2
And then use one or both of the following identities:
1− cos 2𝑣 1+ cos 2𝑣
𝑠𝑖𝑛2 𝑣 = 2
, 𝑐𝑜𝑠 2 𝑣 = 2
To reduce the given integral into an integrable form. The identities are used repeatedly when
m or n or both are greater than 2.
Examples: Evaluate:
1. ∫ 𝑠𝑖𝑛3 4𝑥 𝑐𝑜𝑠 2 4𝑥𝑑𝑥
= ∫ 𝑠𝑖𝑛2 4𝑥𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥
= ∫(1 − 𝑐𝑜𝑠 2 4𝑥)𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥
= ∫(𝑐𝑜𝑠 2 4𝑥 − 𝑐𝑜𝑠 4 4𝑥) sin 4𝑥 𝑑𝑥
MATH 104 – CALCULUS II (INTEGRAL CALCULUS RGREPIQUE
Methods of Integration
1. Product of Sines and Cosines
P1: ∫ 2 sin 𝑢 cos 𝑣 𝑑𝑥 = ∫|sin(𝑢 + 𝑣) + sin(𝑢 − 𝑣)| 𝑑𝑥
P2: ∫ 2 cos 𝑢 cos 𝑣 𝑑𝑥 = ∫|cos(𝑢 + 𝑣) + cos(𝑢 − 𝑣)| 𝑑𝑥
P3: ∫ 2 sin 𝑢 sin 𝑣 𝑑𝑥 = ∫|cos(𝑢 − 𝑣) − cos(𝑢 + 𝑣)| 𝑑𝑥
Examples: Evaluate:
1. ∫ cos 6𝑥 cos 2𝑥 𝑑𝑥
1
= ∫ 2 cos 6𝑥 cos 2𝑥 𝑑𝑥
2
1
= 2 ∫|cos(6𝑥 + 2𝑥) + cos(6𝑥 − 2𝑥)| 𝑑𝑥
1
= 2
∫|cos 8𝑥 + cos 4𝑥| 𝑑𝑥
1 1
Let u = 8x, du = 8 dx, nf = 8 ; u = 4x, du = 4 dx, nf = 4
1 1 1
= ( sin 8𝑥 + sin 4𝑥) + 𝐶
2 8 4
𝟏 𝟏
= 𝐬𝐢𝐧 𝟖𝒙 + 𝐬𝐢𝐧 𝟒𝒙 + 𝑪
𝟏𝟔 𝟖
2. ∫ 3 sin 5𝑥 cos 4𝑥 𝑑𝑥
3
= ∫ 2 sin 5𝑥 cos 4𝑥 𝑑𝑥
2
3
= 2 ∫|sin(5𝑥 + 4𝑥) + sin(5𝑥 − 4𝑥)| 𝑑𝑥
3
= ∫|sin 9𝑥 + sin 𝑥| 𝑑𝑥
2
3 1
= (− cos 9𝑥 − cos 𝑥) + 𝐶
2 9
𝟏 𝟑
= − 𝐜𝐨𝐬 𝟗𝒙 − 𝐜𝐨𝐬 𝒙 + 𝑪
𝟔 𝟐
3. ∫ sin(1 − 2𝑥) sin(2𝑥 + 3) 𝑑𝑥 , u = 1 – 2x, v = 2x + 3
1
= ∫ 2 sin(1 − 2𝑥) sin(2𝑥 + 3)𝑑𝑥
2
1
= ∫|𝑐𝑜𝑠[(1 − 2𝑥) − (2𝑥 + 3)] − 𝑐𝑜𝑠[(1 − 2𝑥) + (2𝑥 + 3)]|
2
1
= ∫|cos(−4𝑥 − 2) − cos(4)| 𝑑𝑥
2
1 1
= {− sin(−4𝑥 − 2) − 𝑥𝑐𝑜𝑠 4} + 𝐶
2 4
𝟏 𝟏
= − 𝟖 𝐬𝐢𝐧(−𝟒𝒙 − 𝟐) − 𝟐 𝒙𝒄𝒐𝒔 𝟒 + 𝑪
MATH 104 – CALCULUS II (INTEGRAL CALCULUS RGREPIQUE
, 2
2. Powers of Sines and Cosines
∫ 𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 𝑑𝑥
Where: v = differentiable function of x
m, n = real numbers.
CASE 1: When m is a positive odd integer and n is any number, we may write:
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛𝑚−1 𝑣 𝑐𝑜𝑠 𝑛 𝑣) sin 𝑣
Since m is odd, then m – 1 is even and therefore, we may use the trigonometry identity,
𝑠𝑖𝑛2 𝑣 = 1 − 𝑐𝑜𝑠 2 𝑣
To express 𝑠𝑖𝑛𝑚−1 𝑣 in terms of the powers of cos v. Then the given integral is reduced to the
form,
∫(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 cos 𝑣) sin 𝑣 𝑑𝑥
which can now be evaluated by use of I4 by letting u = cos v.
CASE 2: When m is any number and n is a positive odd number, we may write:
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛−1 𝑣) cos 𝑣
Since m is odd, then m – 1 is even and therefore, we may use the trigonometry identity,
𝑐𝑜𝑠 2 𝑣 = 1 − 𝑠𝑖𝑛2 𝑣
To express 𝑠𝑖𝑛𝑚−1 𝑣 in terms of the powers of cos v. Then the given integral is reduced to the
form,
∫(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 sin 𝑣) cos 𝑣 𝑑𝑥
which can now be evaluated by use of I4 by letting u = sin v.
CASE 3: When m and n are both even integers (either both positive or one positive and one zero),
we may write:
𝑚 𝑛
𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = (𝑠𝑖𝑛2 𝑣 ) 2 (𝑐𝑜𝑠 2 𝑣 ) 2
And then use one or both of the following identities:
1− cos 2𝑣 1+ cos 2𝑣
𝑠𝑖𝑛2 𝑣 = 2
, 𝑐𝑜𝑠 2 𝑣 = 2
To reduce the given integral into an integrable form. The identities are used repeatedly when
m or n or both are greater than 2.
Examples: Evaluate:
1. ∫ 𝑠𝑖𝑛3 4𝑥 𝑐𝑜𝑠 2 4𝑥𝑑𝑥
= ∫ 𝑠𝑖𝑛2 4𝑥𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥
= ∫(1 − 𝑐𝑜𝑠 2 4𝑥)𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥
= ∫(𝑐𝑜𝑠 2 4𝑥 − 𝑐𝑜𝑠 4 4𝑥) sin 4𝑥 𝑑𝑥
MATH 104 – CALCULUS II (INTEGRAL CALCULUS RGREPIQUE
