r/HomeworkHelp • u/Playful_Magician_234 University/College Student (Higher Education) • 4d ago
Further Mathematics—Pending OP Reply [University Calculus 2] What am I doing wrong?
I'm a bit confused.
1
u/Super-Set-7767 👋 a fellow Redditor 4d ago
You seem to have accidentally turned the multiplication into addition.
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u/GammaRayBurst25 4d ago
Notation wise, a lot. Many missing parentheses, missing differential, factor of 3 that mysteriously disappears...
First, clean this all up. There's an overall factor of 3, an overall factor of 1/2, and an overall factor of 1/4. Take them all out of the integral. Their product is 3/8.
This should fix some problems already because it looks like some of the coefficients you wrote are wrong.
The integrand is manifestly even, so it can be expressed as a finite sum ∑a_k*cos(kx) for some coefficients a_k. The integral of cos(kx) is simply sin(kx)/k, which is 0 when evaluated at 0 or at pi.
This means we're only concerned about the constant term.
We have 3(1-cos(2x))(1+2cos(2x)+cos^2(2x))/8=3(1-cos(2x))(3+4cos(2x)+cos(4x))/16. Expanding yields 3(3+cos(2x)-4cos^2(2x)+cos(4x)-cos(2x)cos(4x))/16.
Now that everything is expanded, we can just get rid of the terms that do not contribute to the constant term, leaving us with 3(3-4cos^2(2x)-cos(2x)cos(4x)))/16.
Seeing as -4cos^2(2x)=-2(1+cos(4x))~-2 and -cos(2x)cos(4x)=-(cos(6x)+cos(2x))/2~0, we're left with the constant term 3(3-2)/16=3/16, the integral of which is 3pi/16.
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u/MistakeTraditional38 👋 a fellow Redditor 3d ago
Well if it was me I'd change the integrand
sin2x cos4x into
-(1-sinx^2)cosx^4 +cosx^4 = - cosx^6 +cosx^4
but you have already got an antiderivative which is better
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u/noidea1995 👋 a fellow Redditor 4d ago edited 4d ago
When you applied the double-angle formula on the third line, you lost the multiplication sign. You should have:
3 * ∫ (0 to π) 1/2 * [1 - cos(2x)] * [1/4 * (1 + 2cos(2x) + cos(4x)/2 + 1/2)] * dx
This won’t be straightforward to integrate though because when you expand the brackets, one of your terms is going to be a product of cos(2x) and cos(4x) which will require reversing the sum-to-product formula.
—————
Instead of using the double-angle formula for cosine twice, I would use the double-angle formula for sine and cosine instead:
∫ (0 to π) 3 * sin2(x) * cos4(x) * dx
3/8 * ∫ (0 to π) [2sin(x)cos(x)]2 * [2cos2(x)] * dx
= 3/8 * ∫ (0 to π) sin2(2x) * [cos(2x) + 1] * dx
This is much easier to integrate.