How do you evaluate the integral #int cos(root3x)#?

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Answer 1 · 2017-04-02T18:22:05

#intcos(root3x)dx=3root3(x^2)sin(root3x)+6root3xcos(root3x)-6sin(root3x)+C#

#I=intcos(root3x)dx#

First, let #t=root3x#. This implies that #t^3=x#, which tells us that #3t^2dt=dx#. We can substitute these in:

#I=intcos(t)(3t^2)dt=int3t^2cos(t)dt#

Now we perform integration by parts. Let:

#{(u=3t^2,=>,du=6tdt),(dv=cos(t)dt,=>,v=sin(t)):}#

Then:

#I=3t^2sin(t)-int6tsin(t)dt#

Integration by parts again:

#{(u=6t,=>,du=6dt),(dv=sin(t)dt,=>,v=-cos(t)):}#

Paying attention to sign:

#I=3t^2sin(t)-(6t(-cos(t))-int(-6cos(t))dt)#

#I=3t^2sin(t)+6tcos(t)-int6cos(t)dt#

#I=3t^2sin(t)+6tcos(t)-6sin(t)+C#

Using #t=root3x#:

#I=3root3(x^2)sin(root3x)+6root3xcos(root3x)-6sin(root3x)+C#