What is the integral of #int (sin(2x)*sin(5x)dx)#?

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Answer 1 · 2017-02-27T07:46:25

The answer is #==1/6sin3x-1/14sin7x+C#

We use,

#cos(a-b)=cosacosb+sinasinb#

#cos(a+b)=cosacosb-sina sinb#

#cos(a-b)-cos(a+b)=2sinasinb#

Here we have,

#a=5x#

#b=2x#

#2sin5xsin2x=cos3x-cos7x#

Therefore,

#intsin5xsin2xdx=1/2intcos3dx-1/2intcos7xdx#

#=1/2(sin3x)/3-1/2(sin7x)/7+C#

#=1/6sin3x-1/14sin7x+C#