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

Question

1602 views around the world

You can reuse this answer
Creative Commons License

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 $t=root3x$ :

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

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