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

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 (A − B)], 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) = 1/2⋅1/5⋅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 (A − B)], 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 (A − B)], 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 (A − B)], with A=ωt 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 (A − B)], 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 (A − B)], with A = B = x, which reduces to:

sin2 (x) = −(1⁄2)(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) = −(1⁄2)(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) = (1⁄2)(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) = (1⁄2) (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)

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