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,

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

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

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

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