Integrate (sinx-5cosx)/(sinx+cosx)

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• March 2nd 2013, 09:38 PM
quanta123
Integrate (sinx-5cosx)/(sinx+cosx)
I'm not sure how to integrate this function. I tried converting the numerator into a(sinx+cosx)+b(cosx-sinx), but I couldn't find anything that worked. It didn't really simplify the integrand.
• March 2nd 2013, 10:36 PM
Soroban
Re: Integrate (sinx-5cosx)/(sinx+cosx)
Hello, quanta123!

I found a way, but it's pretty elaborate.
I'm sure there is a simpler way.

I used these identities:
. . . . . $\sin^2x \:=\:\frac{1-\cos2x}{2}$
. . . . . $\cos^2x \:=\:\frac{1+\cos2x}{2}$
n $2\sin x\cos x \:=\:\sin2x$

$\cos^2\!x - \sin^2\!x \:=\:\cos2x$

Quote:

$\text{Integrate: }\:\int\frac{\sin x - 5\cos x}{\sin x - \cos x}\,dx$

Multiply by $\frac{\sin x + \cos x}{\sin x + \cos x}:$

. . $\frac{\sin x - 5\cos x}{\sin x - \cos x}\cdot\frac{\sin x+\cos x}{\sin x+\cos x} \;=\; \frac{\sin^2\!x + \sin x\cos x - 5\sin x\cos x - 5\cos^2\!x}{\sin^2\!x-\cos^2\!x}$

. . $=\;\frac{\sin^2\!x - 4\sin x\cos x - 5\cos^2\!x}{-(\cos^2\!x - \sin^2\!x)} \;=\;\frac{\frac{1-\cos2x}{2} - 2\sin2x - 5\left(\frac{1+\cos2x}{2}\right)}{-\cos2x}$

. . $=\;\frac{1-\cos2x - 4\sin2x -5-5\cos2x}{-2\cos2x} \;=\;\frac{-4-4\sin2x - 6\cos2x}{-2\cos2x}$

. . $=\;\frac{2 + 2\sin2x + 3\cos2x}{\cos2x} \;=\; \frac{2}{\cos2x} + \frac{2\sin2x}{\cos2x} + \frac{3\cos2x}{\cos2x}$

. . $=\;2\sec2x + 2\tan2x + 3$

Got it?
• March 3rd 2013, 03:40 AM
ibdutt
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Re: Integrate (sinx-5cosx)/(sinx+cosx)
• March 3rd 2013, 06:29 AM
Esteban
Re: Integrate (sinx-5cosx)/(sinx+cosx)
I think there's a simplest way...

$\displaystyle \int \dfrac{\sin x -5\cos x}{\sin x -\cos x}dx=\int \dfrac{\sin x -cos x -4\cos x}{\sin x - \cos x}dx=\int dx-4\int \dfrac{\cos x}{\sin x -\cos x}dx$

and for the last one integral, using $t=\tan \frac{x}{2}$ will be using simple algebra with a rectangle triangle with $t,1,\sqrt{t^2+1}$ sides
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