I would like to solve the ∫sin^2(pix) dx
Using the given substitution identity: sin^2(x) = (1/2)(1-cos2x)
This is what I did so far:
∫sin^2(pix) let u = pix
du = pi dx
(1/pi)∫sin^2(u)du
Applying the identity is where I'm lost on how to continue further..
(1/pi)∫(1/2)(1-cos2u?) du or is it cos2x?
I went further with cos2u by doing:
(1/2pi)∫(1-cos2u) du
(1/2pi)(x - (sin(2pix)/2pi)) +C
However, I'm pretty sure I messed up on the cos2x part of substituting since that answer doesn't seem correct..
Any help is greatly appreciated!
∫sin^2(π x) dx
= ∫(1/2)(1-cos(2πx) ) dx ---> you had that
= ∫(1/2 - (1/2)(cos 2x) dx
At this point you should be able to put your brain in reverse and ask yourself, "what do I need so that the derivative is 1/2 - (1/2)(cos2πx)
remembering that d(sin kx)/dx = k cos kx
I would get
(1/2)x - (1/4π )sin (2πx) + c
you could even get fancy by recalling that
sin (2πx) = 2(sin πx)(cos πx) , but .....
Thanks! So I was on the right track, but messed up near the ending.
To solve the integral ∫sin^2(pix) dx, you started by using the given substitution identity sin^2(x) = (1/2)(1 - cos2x).
Your initial substitution was correct: let u = pix. Then, du = pi dx.
So, the integral becomes: (1/pi) ∫ sin^2(u) du.
Now, to apply the identity sin^2(x) = (1/2)(1 - cos2x), you substitute it in for sin^2(u):
(1/pi) ∫ (1/2)(1 - cos2u) du.
Next, you can distribute the (1/2) term within the integral:
(1/2pi) ∫ (1 - cos2u) du.
Now, you can integrate term by term:
(1/2pi) [∫ 1 du - ∫ cos2u du].
The integral of 1 du is simply u:
(1/2pi) [u - ∫ cos2u du].
To integrate ∫ cos2u du, you can use the substitution v = 2u, which gives you dv = 2 du:
(1/2pi) [u - (1/2) ∫ cos(v) dv].
Simplifying further, you have:
(1/2pi) [u - (1/2) sin(v)].
Now, substitute back the original variable u = pix and v = 2u:
(1/2pi) [pix - (1/2) sin(2pix)].
Finally, you can simplify the expression further:
(1/2pi) pix - (1/4pi) sin(2pix) + C.
So, the correct solution to the integral ∫sin^2(pix) dx is:
(1/2pi) pix - (1/4pi) sin(2pix) + C.
Double-check your calculations with this solution to ensure accuracy.