How to integrate cos⁡(7x)−cos⁡(8x)1+2cos⁡(5x)

Question

How to integrate   cos⁡(7x)−cos⁡(8x)1+2cos⁡(5x)

puhnut1m · Accepted Answer

The rule is to multiply above and below by the

\sin (5 x)

here

\frac{\sin (5 x) (\cos (7 x) - \cos (8 x))}{\sin (5 x) + \sin (10 x)}

= \frac{\sin (5 x) 2 \sin (15 \frac{x}{2}) \sin (\frac{x}{2})}{2 \sin (15 \frac{x}{2}) \cos (5 \frac{x}{2})}

= 2 \sin (5 \frac{x}{2}) \sin (\frac{x}{2})

Now its easy integration right?

kaluitagf · Accepted Answer

HINT:  cos⁡y−cos⁡(6A+y)1+2cos⁡2A=2sin⁡(3A+y)3sin⁡3A1+2(1−2sin2A)  =2sin⁡(3A+y)sin⁡A(3−4sin2A)3−4sin2A=cos⁡(2A+y)−cos⁡(4A+y)  Here 2A=5x, 6A+y=8x⇒y=−7x

karton · Accepted Answer

Let I=∫cos⁡7x−cos⁡8x1+2cos⁡5xdx=∫(cos⁡7x+cos⁡3x)−(cos⁡8x+cos⁡2x)−cos⁡3x+cos⁡2x1+2cos⁡5xdxcosC+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)I=∫2cos⁡5xcos⁡2x+cos⁡2x−2cos⁡5xcos⁡3x−cos⁡3x1+2cos⁡5xdxI=∫(2cos⁡5x+1)(cos⁡2x−cos⁡3x)1+2cos⁡5xdx=∫(cos⁡2x−cos⁡3x)dxSoI=sin⁡2x2−sin⁡3x3+C

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

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