Asked by Rox
∫ sin(x)sin(3x)dx
find the integral
find the integral
Answers
Answered by
Steve
one way:
sin(3x) = sinx cos2x + cosx sin2x
= sinx(2cos^2(x)-1) + 2cosx sinx cosx
= cos^2(x) sinx - sinx + 2cos^2(x) sinx
= 3cos^2(x)sinx - sinx
∫ = -cos^3(x) + cos(x)
another way:
sinx sin3x = 1/2 cos(2x) - 1/2 cos(4x)
∫ = 1/4 sin2x - 1/8 sin4x
a little manipulation of that also yields
sin^3(x) - cos(x)
the various expressions are not identical, but differ only by a constant C
sin(3x) = sinx cos2x + cosx sin2x
= sinx(2cos^2(x)-1) + 2cosx sinx cosx
= cos^2(x) sinx - sinx + 2cos^2(x) sinx
= 3cos^2(x)sinx - sinx
∫ = -cos^3(x) + cos(x)
another way:
sinx sin3x = 1/2 cos(2x) - 1/2 cos(4x)
∫ = 1/4 sin2x - 1/8 sin4x
a little manipulation of that also yields
sin^3(x) - cos(x)
the various expressions are not identical, but differ only by a constant C
Answered by
1
SUIIII
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