Ernest Ryland

2022-01-03

How to integrate
$\frac{\mathrm{cos}\left(7x\right)-\mathrm{cos}\left(8x\right)}{1+2\mathrm{cos}\left(5x\right)}$

puhnut1m

The rule is to multiply above and below by the $\mathrm{sin}\left(5x\right)$ here
$\frac{\mathrm{sin}\left(5x\right)\left(\mathrm{cos}\left(7x\right)-\mathrm{cos}\left(8x\right)\right)}{\mathrm{sin}\left(5x\right)+\mathrm{sin}\left(10x\right)}$
$=\frac{\mathrm{sin}\left(5x\right)2\mathrm{sin}\left(15\frac{x}{2}\right)\mathrm{sin}\left(\frac{x}{2}\right)}{2\mathrm{sin}\left(15\frac{x}{2}\right)\mathrm{cos}\left(5\frac{x}{2}\right)}$
$=2\mathrm{sin}\left(5\frac{x}{2}\right)\mathrm{sin}\left(\frac{x}{2}\right)$
Now its easy integration right?

kaluitagf

HINT:
$\frac{\mathrm{cos}y-\mathrm{cos}\left(6A+y\right)}{1+2\mathrm{cos}2A}=\frac{2\mathrm{sin}\left(3A+y\right)3\mathrm{sin}3A}{1+2\left(1-2{\mathrm{sin}}^{2}A\right)}$
$=\frac{2\mathrm{sin}\left(3A+y\right)\mathrm{sin}A\left(3-4{\mathrm{sin}}^{2}A\right)}{3-4{\mathrm{sin}}^{2}A}=\mathrm{cos}\left(2A+y\right)-\mathrm{cos}\left(4A+y\right)$
Here

karton

Let
$I=\int \frac{\mathrm{cos}7x-\mathrm{cos}8x}{1+2\mathrm{cos}5x}dx=\int \frac{\left(\mathrm{cos}7x+\mathrm{cos}3x\right)-\left(\mathrm{cos}8x+\mathrm{cos}2x\right)-\mathrm{cos}3x+\mathrm{cos}2x}{1+2\mathrm{cos}5x}dx\phantom{\rule{0ex}{0ex}}cosC+\mathrm{cos}D=2\mathrm{cos}\left(\frac{C+D}{2}\right)\mathrm{cos}\left(\frac{C-D}{2}\right)\phantom{\rule{0ex}{0ex}}I=\int \frac{2\mathrm{cos}5x\mathrm{cos}2x+\mathrm{cos}2x-2\mathrm{cos}5x\mathrm{cos}3x-\mathrm{cos}3x}{1+2\mathrm{cos}5x}dx\phantom{\rule{0ex}{0ex}}I=\int \frac{\left(2\mathrm{cos}5x+1\right)\left(\mathrm{cos}2x-\mathrm{cos}3x\right)}{1+2\mathrm{cos}5x}dx=\int \left(\mathrm{cos}2x-\mathrm{cos}3x\right)dx\phantom{\rule{0ex}{0ex}}So\phantom{\rule{0ex}{0ex}}I=\frac{\mathrm{sin}2x}{2}-\frac{\mathrm{sin}3x}{3}+C$

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