anonymous
  • anonymous
\[\int\limits_{}^{}(\cos x- \cos 2x) dx/1-cosx\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is it \[\int\limits_{}^{}{\cos x-\cos 2x \over 1-\cos x}dx?\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
the major problem is to simplify the trig function integration would be easy peasy.

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anonymous
  • anonymous
Yeah, synthetic division to simplify the expression to: \[\cos x+{\cos^2 x-\cos2x \over 1-\cos x}\] But, cos^2x-cos2x=sin^2x. So, it can simplified to: \\[[\cos x+{\sin^2x \over 1-\cos x} =\cos x+{(1-\cos x)(1+\cos x) \over 1- \cos x}=2\cos x+1\]x\] Easy peasy :)
anonymous
  • anonymous
cos x - cos2x = cos x - (2cos^2 x -1) = cosx - cos^2x - cos^2x +1 = cos x(1-cos x) - (1+cos x)(1-cos x) = (1-cos x)(cos x - 1 -cos x) = -(1-cos x)
anonymous
  • anonymous
\[\int\limits_{}^{}(2\cos x+1)dx=2\sin x+x+c\]
anonymous
  • anonymous
Is this the right answer?
anonymous
  • anonymous
yea it is.. thanx man=)
anonymous
  • anonymous
You're welcome!!
anonymous
  • anonymous
wait hold on.. the sign of sin should be negative
anonymous
  • anonymous
I saw a problem you posted earlier that contains a^x, I couldn't solve. Are you sure about the problem?
anonymous
  • anonymous
yea..if you can trust the book..lol
anonymous
  • anonymous
i made some mistake in simplifying the numerator, but it simplifies to (1-cos x)(1+2cos x)
anonymous
  • anonymous
so your integral is : \[\int\limits_{}^{}(1+2cosx) dx\]

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