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klimenkov

  • 3 years ago

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

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  1. SheldonEinstein
    • 3 years ago
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    multiply both sides by cos x + sin x

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

  3. SheldonEinstein
    • 3 years ago
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    Simplify n integrate.

  4. amriju
    • 3 years ago
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    multiply num and den. by cosx + sinx...and solve for sec2x+tan2X

  5. hartnn
    • 3 years ago
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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.....

  6. klimenkov
    • 3 years ago
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    Very nice @hartnn. Your way is the quickest.

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

  8. SheldonEinstein
    • 3 years ago
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    oh sorry there is mod instead of brackets

  9. amriju
    • 3 years ago
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    thats wrong...u should nt have played with our prestige in public...:P @hartnn

  10. hartnn
    • 3 years ago
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    with a negative sign

  11. SheldonEinstein
    • 3 years ago
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    why -ve ?

  12. amriju
    • 3 years ago
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    yeah..ofcrse

  13. hartnn
    • 3 years ago
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    d/dx(cos x-sin x) = -sin x-cos x

  14. SheldonEinstein
    • 3 years ago
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    Oh ok got it :)

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