How do you find the indefinite integral of #int sin(5x)sin(8x)dx#?

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Answer 1 · 2015-09-23T01:28:35

#1/6sin3x - 1/26sin13x+C#

#sinxsiny=1/2(cos(x-y)-cos(x+y))#

#I=int sin5x sin8x dx=1/2 int (cos(-3x) - cos(13x)) dx#

#I=1/2 int cos3xdx - 1/2 int cos13x dx#

#I= 1/6sin3x - 1/26sin13x+C#

Answer 2 · 2015-09-23T01:38:30

Let's ignore the integral symbols for now.

#=> sin(5x)sin(8x)#

It'd be great if we had an identity for this. Maybe we do!

#sinusinv = ?#

Recall that an addition/subtraction identity with #cos(upmv)# contains #sinusinv#.

#cos(u - v) = cosucosv + sinusinv#
#cos(u + v) = cosucosv - sinusinv#

We can acquire #2sinusinv# from this by subtracting the two equations (and thus subtracting #-sinusinv# from #sinusinv#):

#cos(u - v) - cos(u + v)#
#= [cosucosv + sinusinv] - [cosucosv - sinusinv]#

#= 2sinusinv#

Therefore:

#color(green)(sinusinv = (cos(u-v) - cos(u + v))/2)#

Applying this identity, we get:

#= [cos(5x - 8x) - cos(5x + 8x)]/2#

#= [cos(-3x) - cos(13x)]/2#

but #cos(3x) = cos(-3x)#, thus:

#= [cos(3x) - cos(13x)]/2#

This is much easier to do! Bringing back the integral symbols:

#= 1/2int cos(3x)dx - 1/2int cos(13x)dx#

#= 1/2*1/3sin(3x) - 1/2*1/13sin(13x)#

#= color(blue)(1/6sin(3x) - 1/26sin(13x) + C)#