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

Answer 1

Christopher Agresta

#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#

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Answer 2

Christopher Allers

To evaluate the integral ∫ cos(√3x) dx, you can use a substitution method. Let u = √3x, then du = √3 dx. Solving for dx, you get dx = du/√3. Substitute these into the integral:

∫ cos(√3x) dx = ∫ cos(u) * (du/√3) = (1/√3) ∫ cos(u) du

Now, integrate cos(u) with respect to u:

(1/√3) ∫ cos(u) du = (1/√3) * sin(u) + C

Substitute back u = √3x:

(1/√3) * sin(√3x) + C

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.