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anonymous

  • 5 years ago

\[\int\limits_{}^{}(\cos x- \cos 2x) dx/1-cosx\]

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  1. anonymous
    • 5 years ago
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    Is it \[\int\limits_{}^{}{\cos x-\cos 2x \over 1-\cos x}dx?\]

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    the major problem is to simplify the trig function integration would be easy peasy.

  4. anonymous
    • 5 years ago
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    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 :)

  5. anonymous
    • 5 years ago
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    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)

  6. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}(2\cos x+1)dx=2\sin x+x+c\]

  7. anonymous
    • 5 years ago
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    Is this the right answer?

  8. anonymous
    • 5 years ago
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    yea it is.. thanx man=)

  9. anonymous
    • 5 years ago
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    You're welcome!!

  10. anonymous
    • 5 years ago
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    wait hold on.. the sign of sin should be negative

  11. anonymous
    • 5 years ago
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    I saw a problem you posted earlier that contains a^x, I couldn't solve. Are you sure about the problem?

  12. anonymous
    • 5 years ago
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    yea..if you can trust the book..lol

  13. anonymous
    • 5 years ago
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    i made some mistake in simplifying the numerator, but it simplifies to (1-cos x)(1+2cos x)

  14. anonymous
    • 5 years ago
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    so your integral is : \[\int\limits_{}^{}(1+2cosx) dx\]

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