xapproachesinfinity
  • xapproachesinfinity
any easy way to do this than half angle formulas \[\int_0^\pi\sin^4\theta \cos^2 \theta d\theta \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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xapproachesinfinity
  • xapproachesinfinity
the half angle formulas involve a lot of work, so i was wondering about another method
SolomonZelman
  • SolomonZelman
I tried sin^4■ - sin^6■, but sin^6■ is too long to work out...
SolomonZelman
  • SolomonZelman
let's see if I can find something good and quick

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xapproachesinfinity
  • xapproachesinfinity
yeah too long steps, i got pretty tired lol
xapproachesinfinity
  • xapproachesinfinity
your aproach is the first thing i attempted then i moved to using half angles from the very first step but seeems really long
xapproachesinfinity
  • xapproachesinfinity
i think i will just try that first one again and hope i don't make a mistake
SolomonZelman
  • SolomonZelman
it seems like one of the "rather long" problems....
xapproachesinfinity
  • xapproachesinfinity
i hate long ones i give up easily lol
anonymous
  • anonymous
use those silly reduction formulas found in the back of the book
SolomonZelman
  • SolomonZelman
everything that I do is becoming too long for a simple problem
SolomonZelman
  • SolomonZelman
I haven't ever used them ...
xapproachesinfinity
  • xapproachesinfinity
neither do I, i think i will just go with that lol to reduce the trouble
SolomonZelman
  • SolomonZelman
Yeah, I suppose.... (I mean I could solve it just by u-substitution in a year or two, but it isn't worth it, also I am doing another problem on my own time)
xapproachesinfinity
  • xapproachesinfinity
did over 60 problems my head is burning now lol there were many who were long
SolomonZelman
  • SolomonZelman
(It is Σ 1/ln(Γ(kx)) for different k .... ) I will see if there are some open questions to do, but, I never liked integrals with a happy end but someone dies (integratable, but too long).
SolomonZelman
  • SolomonZelman
using series for these integrals would be in a snap, but series isn't like really a complete answer though.
SolomonZelman
  • SolomonZelman
In any case, good luck!! cu
ParthKohli
  • ParthKohli
\[u = \tan x\]is a better substitution here, as I recall.
ParthKohli
  • ParthKohli
ok no
SolomonZelman
  • SolomonZelman
that looked scary...
anonymous
  • anonymous
I don't know about easy, but if you're really intent on going around using the half-angle formulas, you can try various substitutions that exploit symmetry to rewrite the initial integral. \[\mathcal{I}=\int_0^\pi\sin^4t\cos^2t\,\mathrm{d}t\]Replacing \(t\) with \(\pi-t\), it's easy to see that \[\mathcal{I}=\int_0^\pi\cos^4t\sin^2t\,\mathrm{d}t\]Furthermore, replacing \(t\) with \(\dfrac{\pi}{2}-t\) gives \[\mathcal{I}=\int_{-\pi/2}^{\pi/2}\sin^4t\cos^2t\,\mathrm{d}t=\int_{-\pi/2}^{\pi/2}\cos^4t\sin^2t\,\mathrm{d}t\]Since both integrands are even, you get \[\frac{1}{2}\mathcal{I}=\int_0^{\pi/2}\sin^4t\cos^2t\,\mathrm{d}t=\int_0^{\pi/2}\cos^4t\sin^2t\,\mathrm{d}t\]which further tells you that \[\mathcal{I}=\int_0^{\pi/2}(\sin^4t+\cos^4t)\,\mathrm{d}t\]What can you do with this?
anonymous
  • anonymous
DeMoivre's theorem could work in "simplifying" the current integrand. \[\begin{align*}\Re\left\{e^{4it}\right\}&=\Re\left\{(\cos t+i\sin t)^4\right\}\\[1ex] &=\cos^4t+\sin^4t-\cos^2t\sin^2t\\[1ex] &=\cos4t\end{align*}\]So you have \[\mathcal{I}=\int_0^{\pi/2}\left(\cos4t+\cos^2t\sin^2t\right)\,\mathrm{d}t\]
anonymous
  • anonymous
From here, \[\cos^2t\sin^2t=\left(\frac{1}{2}\sin2t\right)^2=\frac{1}{4}\sin^22t\]Let \[\mathcal{J}=\int_0^{\pi/2}\sin^22t\,\mathrm{d}t\]so that \[\mathcal{I}=\int_0^{\pi/2}\cos4t\,\mathrm{d}t+\frac{1}{4}\mathcal{J}\]Replacing \(t\) with \(\dfrac{\pi}{4}-t\) and \(\dfrac{\pi}{2}-t\), you can show that \[\mathcal{J}=\int_0^{\pi/2}\sin^22t\,\mathrm{d}t=\int_0^{\pi/2}\cos^22t\,\mathrm{d}t\]so that \[\mathcal{J}=\frac{1}{2}\int_0^{\pi/2}\mathrm{d}t\]
thomas5267
  • thomas5267
That's even more work than the half-angle...
xapproachesinfinity
  • xapproachesinfinity
interesting approach @SithsAndGiggles

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