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klimenkov Group Title

What about this? \[\int{\frac{\sin x+\cos x}{\cos x - \sin x}}dx\]

  • one year ago
  • one year ago

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  1. SheldonEinstein Group Title
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    multiply both sides by cos x + sin x

    • one year ago
  2. SheldonEinstein Group Title
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    You will get like this : \[\large{\frac{1 + 2cosx.sinx}{cos^2x-sin^2x}}\]

    • one year ago
  3. SheldonEinstein Group Title
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    Simplify n integrate.

    • one year ago
  4. amriju Group Title
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    multiply num and den. by cosx + sinx...and solve for sec2x+tan2X

    • one year ago
  5. hartnn Group Title
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    well, do u know this standard identity \(\huge \int \frac{f'(x)}{f(x)}dx=ln|f(x)|+c\) no need to multiply or divide anything.....

    • one year ago
  6. klimenkov Group Title
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    Very nice @hartnn. Your way is the quickest.

    • one year ago
  7. SheldonEinstein Group Title
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    @hartnn so it will be : \(\large{ln (\cos x - \sin x)+C}\)

    • one year ago
  8. SheldonEinstein Group Title
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    oh sorry there is mod instead of brackets

    • one year ago
  9. amriju Group Title
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    thats wrong...u should nt have played with our prestige in public...:P @hartnn

    • one year ago
  10. hartnn Group Title
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    with a negative sign

    • one year ago
  11. SheldonEinstein Group Title
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    why -ve ?

    • one year ago
  12. amriju Group Title
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    yeah..ofcrse

    • one year ago
  13. hartnn Group Title
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    d/dx(cos x-sin x) = -sin x-cos x

    • one year ago
  14. SheldonEinstein Group Title
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    Oh ok got it :)

    • one year ago
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