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Evaluate sinxsin2xsin3xdx

Answer
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Hint: Here as you can see the equation is in the form of sinAsinB, so we apply the formula and then simplify the integral.

Complete step-by-step answer:
As you know
sinAsinB=12(cos(AB)cos(A+B))
Applying this, we get
12(cos(x2x)cos(x+2x))sin3xdx
We know that cos(θ)=cos(θ)
12(cosxcos3x)sin3xdx
Now break the integration
12(cosxsin3x)dx12(cos3xsin3x)dx
As you know,
cosAsinB=12(sin(B+A)+sin(BA)) , and sinAcosA=12sin2A
Applying this, we get
12×12(sin(3x+x)+sin(3xx))dx12×12(sin6x)dx
Now again break the integrals
14sin4xdx+14sin2xdx14(sin6x)dx
Now apply integration
As we know sin(ax) integration is cos(ax)a
14(cos4x4+cos2x2cos6x6) + C
14(cos4x4cos2x2+cos6x6) + C
148(3cos4x6cos2x+2cos6x) + C
So, this is your required answer.

Note: In this type of question first apply the trigonometry formula, then simplify the integration you will get your answer. It can also be solved by applying integration by parts but it will be tedious.
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