integrate sinx/sin4x

**ice_syncer** · November 13th 2010, 12:36 AM

Integrate

Code:

sinx/sin4x

I exapned sin4x and did some partial fractions stuff, but the answer which I get is not the back answer

**mr fantastic** · November 13th 2010, 12:53 AM

Originally Posted by ice_syncer

Integrate

Code:

sinx/sin4x

I exapned sin4x and did some partial fractions stuff, but the answer which I get is not the back answer

integrate Sin[x]/Sin[4x] - Wolfram|Alpha

Click on Show steps.

**ice_syncer** · November 13th 2010, 01:34 AM

;D

**tonio** · November 13th 2010, 03:49 AM

Originally Posted by mr fantastic

integrate Sin[x]/Sin[4x] - Wolfram|Alpha

Click on Show steps.

Wolfram's steps look too frightening. I'd rather go for a trigonometric substitution:

$\displaystyle{\tan\frac{x}{2}=t\Longrightarrow dx=\frac{2}{1+t^2}\,dt\,,\,\,\cos x=\frac{1-t^2}{1+t^2}\,,\,\,\sin x=\frac{2t}{1+t^2}}$ , so with a little trigonometry:

$\displaystyle{\int \frac{\sin x}{\sin 4x}\,dx=\int\frac{\sin x}{2\sin 2x\cos 2x}\,dx=\int\frac{dx}{4\cos x(2\cos^2x-1)}\,dx}=$

$\displaystyle{\frac{1}{4}\int\frac{(1+t^2)^2}{(1-t^2)\left[t^2-(3+2\sqrt{2})\right]\left[t^2-(3-2\sqrt{2})\right]}}$ , and we have an integral of a rational function.

Indeed this still is a nasty integral, but perhaps a little less messier than in the steps of the proposed solution

in Wolfram's...perhaps.

Tonio

**Krizalid** · November 13th 2010, 05:47 AM

Originally Posted by tonio

$\int\frac{dx}{4\cos x(2\cos^2x-1)}\,dx}$

$\displaystyle\frac{1}{\cos x(2{{\cos }^{2}}x-1)}=\frac{\cos x}{(1-{{\sin }^{2}}x)(1-2{{\sin }^{2}}x)},$ substitute $t=\sin t$ and get

$\begin{aligned} \int{\frac{dt}{(1-{{t}^{2}})(1-2{{t}^{2}})}}&=\int{\frac{2(1-{{t}^{2}})-(1-2{{t}^{2}})}{(1-{{t}^{2}})(1-2{{t}^{2}})}\,dt} \\ & =2\int{\frac{dt}{1-2{{t}^{2}}}}-\int{\frac{dt}{1-{{t}^{2}}}\,dt}, \\ \end{aligned}$

kill those in the same fashion.

-----

that's one of the biggest reasons on why i don't absolute recommend wolfram for tricky problems, we gotta let our people think.

**TheCoffeeMachine** · November 13th 2010, 08:00 AM

Wolfram also made mess of the partial fractions:

If letting $u = \sin{x}$ gives $\displaystyle \int\frac{1}{8 u^4-12 u^2+4}\;{du}$ , then simply write:

$\displaystyle 8u^4-12u^2+4 = 8t^2-12t+4 = 4(2t^2-3t+1)[/Math] [LaTeX ERROR: Convert failed] $ [Math]\displaystyle = \frac{1}{8}\int\frac{1}{(u-1)(2u^2-1)}\;{du}-\frac{1}{8}\int\frac{1}{(u+1)(2u^2-1)}\;{du}$

[LaTeX ERROR: Code too long, max. 1000 characters] $ [LaTeX ERROR: Convert failed] $ [LaTeX ERROR: Convert failed] $ [LaTeX ERROR: Convert failed] $ [LaTeX ERROR: Convert failed] $, we have: [Math]\displaystyle I = \frac{1}{16(1-\sqrt{2})}\ln\left\{\frac{u-1}{\sqrt{2}u-1}\right\}+\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac {u-1}{\sqrt{2}u+1}\right\}$

$\displaystyle -\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{u+1}{\sqrt {2}u-1}\right\}+\frac{1}{16(\sqrt{2}-1)}\ln\left\{\frac{u+1}{\sqrt{2}u+1}\right\}+k$

Thus our original integral evaluates to:

$\displaystyle I = \frac{1}{16(1-\sqrt{2})}\ln\left\{\frac{\sin{x}-1}{\sqrt{2}\sin{x}-1}\right\}+\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac {\sin{x}-1}{\sqrt{2}\sin{x}+1}\right\}$

$\displaystyle -\frac{1}{16(1+\sqrt{2})}\ln\left\{\frac{\sin{x}+1} {\sqrt{2}\sin{x}-1}\right\}+\frac{1}{16(\sqrt{2}-1)}\ln\left\{\frac{\sin{x}+1}{\sqrt{2}\sin{x}+1}\r ight\}+k$

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