Integration of Trignometric Identities |
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Integrals of the form:
and similar ones with products like sin (nx) cos (mx), can be solved by making use of the following trigonometric identities.
Using these identities, such products are expressed as the sum of trigonometric functions. Such sums are generally more straightforward to integrate.
Perform the following integrations.
Click on the questions to reveal their solution
Exercise 1:
Solution:
Use cos (A) cos (B) = 12[cos (A + B) + cos (A − B)], with A = 3x and B = 2x:
∫ cos (3x) cos (2x) dx | = | 12 ∫ ( cos (3x + 2x) + cos (3x − 2x) ) dx |
= | 12 ∫ ( cos (5x) + cos (x) ) dx |
Each term in the integration sign is a function of a linear function of x, i.e.:
Thus:
Exercise 2:
Solution:
Use sin (A) cos (B) = 12[sin (A + B) + sin (A − B)], with A = 5x and B = 3x:
∫ sin (5x) cos (3x) dx | = | 12 ∫ ( sin (5x + 3x) + sin (5x − 3x) ) dx |
= | 12 ∫ ( sin (8x) + sin (2x) ) dx | |
= | − 12⋅18⋅ cos (8x) − 12⋅12⋅ cos (2x) + C | |
= | − 116 cos (8x) − 14 cos (2x) + C |
Exercise 3:
Solution:
Use sin (A) sin (B) = −12[cos (A + B)−cos (A − B)], with A = 6x and B = 4x:
∫ sin (6x) sin (4x) dx | = | −12 ∫ ( cos (6x + 4x) − cos (6x − 4x) ) dx |
= | −12 ∫ ( cos (10x) − cos (2x) ) dx | |
= | − 12⋅110⋅ sin (10x) + 12⋅12⋅ sin (2x) + C | |
= | − 120 sin (10x) + 14 sin (2x) + C |
Exercise 4:
Solution:
Use sin (A) cos (B) = 12[sin (A + B) + sin (A − B)], with A=ωt and B=2ωt:
∫ cos (2ωt) sin (ωt) dt | = | 12 ∫ ( sin (ωt + 2ωt) + sin (ωt − 2ωt) ) dx |
= | 12 ∫ ( sin (3ωt) + sin (−ωt) ) dx | |
= | 12 ∫ ( sin (3ωt) − sin (ωt) ) dx | |
= | − 12⋅13ω⋅ cos (3ωt) + 12⋅1ω⋅ cos (ωt) + C | |
= | − 16ω cos (3ωt) + 12ω cos (ωt) + C |
Exercise 5:
Solution:
Use cos (A) cos (B) = 12[cos (A + B) + cos (A − B)], with A = 4ωt and B = 2ωt:
∫ cos (4ωt) cos (2ωx) dt | = | 12 ∫ ( cos (4ωt+2ωt) + cos (4ωt−2ωt) ) dx |
= | 12 ∫ ( cos (6ωt) + cos (2ωt) ) dx | |
= | 12⋅16ω⋅ sin (6ωt) + 12⋅12ω⋅ sin (2ωt) + C | |
= | 112ω sin (6ωt) + 14ω sin (2ωt) + C |
Exercise 6:
Solution:
Use sin (A) sin (B) = −12[cos (A + B) − cos (A − B)], with A = B = x, which reduces to:
∫ sin2 (x) dx | = | −12 ∫ ( cos (2x) − 1 ) dx |
= | − 12⋅12⋅ sin (2x) + 12x + C | |
= | − 14 sin (2x) + 12x + C |
Exercise 7:
Solution:
As in Exercise 6, sin2 (ωt) reduces to:
∫ sin2 (ωt) dt | = | −12 ∫ ( cos (2ωt) −1 ) dx |
= | −12⋅12ω⋅ sin (2ωt) + 12t + C | |
= | −14ω sin (2ωt) + 12t + C |
Exercise 8:
Solution:
cos2 (t) reduces to:
∫ cos2 (t) dt | = | 12 ∫ ( cos (2t) + 1 ) dx |
= | 12⋅12⋅ sin (2t) + 12t + C | |
= | 14 sin (2t) + 12t + C |
Exercise 9:
Solution:
cos2 (kx) reduces to:
∫ cos2 (kx) dt | = | 12 ∫ ( cos (2kx) + 1 ) dx |
= | 12⋅12k⋅ sin (2kx) + 12x + C | |
= | 14k sin (2kx) + 12x + C |
ƒ(x) | ∫ƒ(x) dx | ƒ(x) | ∫ƒ(x) dx | |
---|---|---|---|---|
xn | xn+1 n+1 (n ≠ −1) | [g(x)]n g'(x) | [g(x)]n+1 n+1 (n ≠ −1) | |
1x | ln x | g'(x)g(x) | ln g(x) | |
ex | ex | ax | axln a (a > 0) | |
sin x | −cos x | sinh x | cosh x | |
cos x | sin x | cosh x | sinh x | |
tan x | − ln cosx | tanh x | ln cosh x | |
cosec x | ln tan x2 | cosech x | ln tanh x2 | |
sec x | ln sec x + tan x | sec x | 2 tan−1ex | |
sec2 x | tan x | sec2 x | tanh x | |
cot x | ln sin x | cot x | ln sinh x | |
sin2 x | x2 − sin 2x4 | sinh2 x | sinh 2x4 − x2 | |
cos2 x | x2 + sin 2x4 | cosh2 x | sinh 2x4 + x2 | |
1a2 + x2 | 1a tan −1 xa (a > 0) | √a2 + x2 | a22 [ sinh−1( xa ) + x √a2−x2 a2 ] | |
1a2 − x2 | 12a ln a + xa − x (0 < x < a) | √a2−x2 | a22 [ sin−1( xa ) + x √a2−x2 a2 ] | |
1x2 − a2 | 12a ln x − ax + a (x > a > 0) | √x2 − a2 | a22 [−cosh−1( xa ) + x √x2−a2 a2 ] | |
1 √ a2 + x2 | ln x + √a2 + x2 a (a > 0) | |||
1 √ a2 − x2 | sin−1 xa (−a < x < a) | 1 √ x2 − a2 | ln x + √x2 − a2 a (x > a > 0) |