Integral of sin(7x)cos(4x)dx?
1 Answer
Feb 20, 2018
#I=-(1/22cos(11x)+1/6cos(3x))+C#
Explanation:
We want to solve
#I=intsin(7x)cos(4x)dx#
Rewrite the integrand using the trigonometric product identity
#sin(a)cos(b)=1/2(sin(a+b)+sin(a-b))#
Therefore
#I=1/2intsin(7x+4x)+sin(7x-4x)dx#
#=1/2intsin(11x)+sin(3x)dx#
#=1/2intsin(11x)+1/2intsin(3x)dx#
#=-1/2*1/11cos(11x)-1/2*1/3cos(3x)+C#
#=-1/22cos(11x)-1/6cos(3x)+C#
#=-(1/22cos(11x)+1/6cos(3x))+C#
