# 1. Theory

Integrals of the form:

sin (nx) sin (mx)

and similar ones with products like sin (nx) cos (mx), can be solved by making use of the following trigonometric identities.

sin (A) sin (B) = −1/2[cos (A + B)−cos (AB)]
sin (A) cos (B) = 1/2[sin (A + B) + sin (AB)]
cos (A) cos (B) = 1/2[cos (A + B) + cos (AB)]

Using these identities, such products are expressed as the sum of trigonometric functions. Such sums are generally more straightforward to integrate.

# 2. Exercises

Perform the following integrations.

Click on the questions to reveal their solution

Exercise 1:

cos (3x) cos (2x) dx

Solution:

Use cos (A) cos (B) = 1/2[cos (A + B) + cos (AB)], with A = 3x and B = 2x:

 ∫  cos (3x) cos (2x) dx = 1/2   ∫  ( cos (3x + 2x) + cos (3x − 2x) ) dx = 1/2   ∫  ( cos (5x) + cos (x) ) dx

Each term in the integration sign is a function of a linear function of x, i.e.:

ƒ(ax + b) dx = 1/a     ƒ(u)du, where u = ax + b, du = a dx, dx = du/a

Thus:

cos (3x) cos (2x) = 1/21/5⋅sin (5x) + 1/2⋅sin (x) + C = 1/10 sin (5x) + 1/2 sin (x) + C

Exercise 2:

sin (5x) cos (3x) dx

Solution:

Use sin (A) cos (B) = 1/2[sin (A + B) + sin (AB)], with A = 5x and B = 3x:

 ∫  sin (5x) cos (3x) dx = 1/2   ∫  ( sin (5x + 3x) + sin (5x − 3x) ) dx = 1/2   ∫  ( sin (8x) + sin (2x) ) dx = − 1/2⋅1/8⋅ cos (8x) − 1/2⋅1/2⋅ cos (2x) + C = − 1/16 cos (8x) − 1/4 cos (2x) + C

Exercise 3:

sin (6x) sin (4x) dx

Solution:

Use sin (A) sin (B) = −1/2[cos (A + B)−cos (AB)], with A = 6x and B = 4x:

 ∫  sin (6x) sin (4x) dx = −1/2   ∫  ( cos (6x + 4x) − cos (6x − 4x) ) dx = −1/2   ∫  ( cos (10x) − cos (2x) ) dx = − 1/2⋅1/10⋅ sin (10x) + 1/2⋅1/2⋅ sin (2x) + C = − 1/20 sin (10x) + 1/4 sin (2x) + C

Exercise 4:

cos (2ωt) sin (ωx) dt, where ω is a constant.

Solution:

Use sin (A) cos (B) = 1/2[sin (A + B) + sin (AB)], with At and B=2ωt:

 ∫  cos (2ωt) sin (ωt) dt = 1/2   ∫  ( sin (ωt + 2ωt) + sin (ωt − 2ωt) ) dx = 1/2   ∫  ( sin (3ωt) + sin (−ωt) ) dx = 1/2   ∫  ( sin (3ωt) − sin (ωt) ) dx = − 1/2⋅1/3ω⋅ cos (3ωt) + 1/2⋅1/ω⋅ cos (ωt) + C = − 1/6ω cos (3ωt) + 1/2ω cos (ωt) + C

Exercise 5:

cos (4ωt) cos (2ωx) dt, where ω is a constant.

Solution:

Use cos (A) cos (B) = 1/2[cos (A + B) + cos (AB)], with A = 4ωt and B = 2ωt:

 ∫  cos (4ωt) cos (2ωx) dt = 1/2   ∫  ( cos (4ωt+2ωt) + cos (4ωt−2ωt) ) dx = 1/2   ∫  ( cos (6ωt) + cos (2ωt) ) dx = 1/2⋅1/6ω⋅ sin (6ωt) + 1/2⋅1/2ω⋅ sin (2ωt) + C = 1/12ω sin (6ωt) + 1/4ω sin (2ωt) + C

Exercise 6:

sin2 (x) dx

Solution:

Use sin (A) sin (B) = −1/2[cos (A + B) − cos (AB)], with A = B = x, which reduces to:

sin2 (x) = −(12)(cos (2x) − 1)
 ∫  sin2 (x) dx = −1/2   ∫  ( cos (2x) − 1 ) dx = − 1/2⋅1/2⋅ sin (2x) + 1/2x + C = − 1/4 sin (2x) + 1/2x + C

Exercise 7:

sin2 (ωt) dt, where ω is a constant.

Solution:

As in Exercise 6, sin2 (ωt) reduces to:

sin2 (ωt) = −(12)(cos (2ωt) − 1)
 ∫  sin2 (ωt) dt = −1/2   ∫  ( cos (2ωt) −1 ) dx = −1/2⋅1/2ω⋅ sin (2ωt) + 1/2t + C = −1/4ω sin (2ωt) + 1/2t + C

Exercise 8:

cos2 (t) dt

Solution:

cos2 (t) reduces to:

cos2 (t) = (12)(cos (2t) + 1)
 ∫  cos2 (t) dt = 1/2   ∫  ( cos (2t) + 1 ) dx = 1/2⋅1/2⋅ sin (2t) + 1/2t + C = 1/4 sin (2t) + 1/2t + C

Exercise 9:

cos2 (kx) dx, where k is a constant.

Solution:

cos2 (kx) reduces to:

cos2 (kx) = (12) (cos (2kx) + 1)
 ∫  cos2 (kx) dt = 1/2   ∫  ( cos (2kx) + 1 ) dx = 1/2⋅1/2k⋅ sin (2kx) + 1/2x + C = 1/4k sin (2kx) + 1/2x + C

# 3. Standard Integrals

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) 1/x ln x g'(x)/g(x) ln g(x) ex ex ax ax/ln 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  x/2 cosech x ln tanh  x/2 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 x/2 − sin 2x/4 sinh2 x sinh 2x/4 − x/2 cos2 x x/2 + sin 2x/4 cosh2 x sinh 2x/4 + x/2 1/ a2 + x2 1/ a tan −1  x/ a   (a > 0) √a2 + x2 a2/ 2 [ sinh−1( x/ a ) + x √a2−x2 / a2 ] 1/ a2 − x2 1/ 2a ln  a + x/ a − x    (0 < x < a) √a2−x2 a2/ 2 [ sin−1( x/ a ) + x √a2−x2 / a2 ] 1/ x2 − a2 1/ 2a ln  x − a/ x + a   (x > a > 0) √x2 − a2 a2/ 2 [−cosh−1( x/ a ) + x √x2−a2 / a2 ] 1/ √ a2 + x2 ln  x + √a2 + x2 / a   (a > 0) 1/ √ a2 − x2 sin−1  x/ a     (−a < x < a) 1/ √ x2 − a2 ln  x + √x2 − a2 / a   (x > a > 0)