What is the integral of #int sin(3x) * cos(4x) dx#?

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Answer 1 · 2016-12-02T15:11:43

The answer is #=-cos7x/14-cosx/2 +C#

#sin(a+b)=sinacosb+sinbcosa#

#sin(a-b)=sinacosb-sinbcosa#

#sin(a+b)+sin(a-b)=2sinacosb#

#sinacosb=1/2(sin(a+b)+sin(a-b))#

Therefore

#cos3xsin4x=1/2(sin(3x+4x)+sin(4x-3x))#

#=1/2(sin7x+sinx)#

So,

#intcos3xsin4xdx=1/2int(sin7x+sinx)#

#=-1/2((cos7x)/7+cosx)+C#

#=-(cos7x)/14-cosx/2 +C#