justconfused
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- May 3, 2013
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I'm stuck on this trigonometric integral problem:
So I got to this point
and I'm not sure where to go from there.
So I got to this point
and I'm not sure where to go from there.
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Trigonometric Integrals
- Thread starter justconfused
- Start date
You will want to use:
\(\displaystyle \sin^2(x) = \dfrac{1-\cos(2x)}{2}\)
\(\displaystyle \cos^2(x) = \dfrac{1+\cos(2x)}{2}\)
For ex.
\(\displaystyle \sin^6(x) = \left(\sin^2(x)\right)^3 = \left(\dfrac{1-\cos(2x)}{2}\right)^3 = -\dfrac{1}{8}\left(\cos^3(2x) - 3\cos^2(2x) + 3\cos(2x) -1\right)\)
Then the above integrable is doable with:
\(\displaystyle \cos^2(2x) = \dfrac{1}{2}\left(1+\cos(4x)\right)\) : Straight-forward
\(\displaystyle \cos^3(2x) = (1-\sin^2(2x))\cdot \cos(2x)\): \(\displaystyle \dfrac{1}{2}(1-u^2)*u'\)
\(\displaystyle \sin^2(x) = \dfrac{1-\cos(2x)}{2}\)
\(\displaystyle \cos^2(x) = \dfrac{1+\cos(2x)}{2}\)
For ex.
\(\displaystyle \sin^6(x) = \left(\sin^2(x)\right)^3 = \left(\dfrac{1-\cos(2x)}{2}\right)^3 = -\dfrac{1}{8}\left(\cos^3(2x) - 3\cos^2(2x) + 3\cos(2x) -1\right)\)
Then the above integrable is doable with:
\(\displaystyle \cos^2(2x) = \dfrac{1}{2}\left(1+\cos(4x)\right)\) : Straight-forward
\(\displaystyle \cos^3(2x) = (1-\sin^2(2x))\cdot \cos(2x)\): \(\displaystyle \dfrac{1}{2}(1-u^2)*u'\)
Last edited:
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I'm stuck on this trigonometric integral problem:
View attachment 2812
So I got to this point View attachment 2813
and I'm not sure where to go from there.
sin4(x)cos2(x) = sin2(x)*1/4 *[sin(2x)]2 = 1/8 * [1 - cos(2x)] * [sin(2x)]2 = 1/8 *[ sin2(2x) - sin2(2x) * cos(2x)] = 1/16*[1 - cos(4x) - 2* sin2(2x) * cos(2x)]
Hello, justconfused!
We need these two identities:
. . \(\displaystyle \sin^2\!\theta \:=\:\dfrac{1-\cos2\theta}{2}\)
. . \(\displaystyle 2\sin\theta\cos\theta \:=\:\sin2\theta\)
I would approach it like this . . .
\(\displaystyle \displaystyle 3\int (\sin^2\!x)(\sin^2\!x\cos^2\!x)\,dx\)
. . \(\displaystyle \displaystyle =\;3\int\left(\frac{1-\cos 2x}{2}\right)\left(\frac{\sin2x}{2}\right)^2\,dx\)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{8}\int(1-\cos2x)\sin^2\!2x\,dx\)
. . \(\displaystyle =\;\tfrac{3}{8}\int(\sin^2\!2x - \sin^2\!2x\cos2x)\,dx \)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{8}\int\left(\frac{1-\cos4x}{2} - \sin^2\!2x\cos2x\right)\,dx \)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{16}\int\left(1 - \cos4x - 2\sin^2\!2x\cos2x\right)\,dx\)
. . \(\displaystyle \displaystyle =\; \tfrac{3}{16}\left[\int 1\,dx - \int\cos4x\,dx - \int\sin^2\!2x\cos2x\,dx\right] \)
Can you finish it now?
We need these two identities:
. . \(\displaystyle \sin^2\!\theta \:=\:\dfrac{1-\cos2\theta}{2}\)
. . \(\displaystyle 2\sin\theta\cos\theta \:=\:\sin2\theta\)
\(\displaystyle \displaystyle\int 3\sin^4\!x\cos^2\!x\,dx\)
I would approach it like this . . .
\(\displaystyle \displaystyle 3\int (\sin^2\!x)(\sin^2\!x\cos^2\!x)\,dx\)
. . \(\displaystyle \displaystyle =\;3\int\left(\frac{1-\cos 2x}{2}\right)\left(\frac{\sin2x}{2}\right)^2\,dx\)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{8}\int(1-\cos2x)\sin^2\!2x\,dx\)
. . \(\displaystyle =\;\tfrac{3}{8}\int(\sin^2\!2x - \sin^2\!2x\cos2x)\,dx \)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{8}\int\left(\frac{1-\cos4x}{2} - \sin^2\!2x\cos2x\right)\,dx \)
. . \(\displaystyle \displaystyle =\;\tfrac{3}{16}\int\left(1 - \cos4x - 2\sin^2\!2x\cos2x\right)\,dx\)
. . \(\displaystyle \displaystyle =\; \tfrac{3}{16}\left[\int 1\,dx - \int\cos4x\,dx - \int\sin^2\!2x\cos2x\,dx\right] \)
Can you finish it now?