Integratethe following ...........,,....,: sin4x.cos3x.sin2x
Ritz , 6 Years ago
Grade 12
Integral Calculus> Integrate the following ...........,,.......
1 Answers
Arun
Last Activity: 6 Years ago
Note that sin(A)sin(B) = (1/2) [cos(A - B) - cos(A + B)].
Also, cos(-A) = cos(A).
Therefore,
sin(2x)sin(4x) = (1/2) [cos(2x) - cos(6x)].
Also note that cos(A)cos(B) = (1/2) [cos(A + B) + cos(A - B)].
∫sin(2x)cos(3x)sin(4x) dx
= ∫ (1/2) [cos(2x) - cos(6x)] cos(3x) dx
= (1/2) ∫ [cos(2x)cos(3x) - cos(6x)cos(3x)] dx
= (1/2)(1/2) ∫ [cos(5x) + cos(x) - cos(9x) - cos(3x)] dx
= (1/4) ∫[cos(x) - cos(3x) + cos(5x) - cos(9x)] dx
= (1/4) [ sin(x) - sin(3x)/3 + sin(5x)/5 - sin(9x)/9 ]
Also, cos(-A) = cos(A).
Therefore,
sin(2x)sin(4x) = (1/2) [cos(2x) - cos(6x)].
Also note that cos(A)cos(B) = (1/2) [cos(A + B) + cos(A - B)].
∫sin(2x)cos(3x)sin(4x) dx
= ∫ (1/2) [cos(2x) - cos(6x)] cos(3x) dx
= (1/2) ∫ [cos(2x)cos(3x) - cos(6x)cos(3x)] dx
= (1/2)(1/2) ∫ [cos(5x) + cos(x) - cos(9x) - cos(3x)] dx
= (1/4) ∫[cos(x) - cos(3x) + cos(5x) - cos(9x)] dx
= (1/4) [ sin(x) - sin(3x)/3 + sin(5x)/5 - sin(9x)/9 ]
Regards
Arun (askIITians forum expert)
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