Question

any easy way to do this than half angle formulas
\[\int_0^\pi\sin^4\theta \cos^2 \theta d\theta \]

Answer 1

the half angle formulas involve a lot of work, so i was wondering about another method

Answer 2

I tried sin^4■ - sin^6■, but sin^6■ is too long to work out...

Answer 3

let's see if I can find something good and quick

Answer 4

yeah too long steps, i got pretty tired lol

Answer 5

your aproach is the first thing i attempted then i moved to using half angles from the very first step
but seeems really long

Answer 6

i think i will just try that first one again and hope i don't make a mistake

Answer 7

it seems like one of the "rather long" problems....

Answer 8

i hate long ones i give up easily lol

Answer 9

use those silly reduction formulas found in the back of the book

Answer 10

everything that I do is becoming too long for a simple problem

Answer 11

I haven't ever used them ...

Answer 12

neither do I, i think i will just go with that lol
to reduce the trouble

Answer 13

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)

Answer 14

did over 60 problems my head is burning now lol
there were many who were long

Answer 15

(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).

Answer 16

using series for these integrals would be in a snap, but series isn't like really a complete answer though.

Answer 17

In any case, good luck!!
cu

Answer 18

\[u = \tan x\]is a better substitution here, as I recall.

Answer 19

ok no

Answer 20

that looked scary...

Answer 21

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?

Answer 22

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
\[\mathcal{I}=\int_0^{\pi/2}\left(\cos4t+\cos^2t\sin^2t\right)\,\mathrm{d}t\]

Answer 23

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
\[\mathcal{J}=\frac{1}{2}\int_0^{\pi/2}\mathrm{d}t\]

Answer 24

That's even more work than the half-angle...

Answer 25

interesting approach @SithsAndGiggles