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



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SheldonEinstein
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multiply both sides by cos x + sin x

SheldonEinstein
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You will get like this :
\[\large{\frac{1 + 2cosx.sinx}{cos^2xsin^2x}}\]

SheldonEinstein
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Simplify n integrate.

amriju
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multiply num and den. by cosx + sinx...and solve for sec2x+tan2X

hartnn
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well, do u know this standard identity
\(\huge \int \frac{f'(x)}{f(x)}dx=lnf(x)+c\)
no need to multiply or divide anything.....

klimenkov
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Very nice @hartnn. Your way is the quickest.

SheldonEinstein
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@hartnn so it will be : \(\large{ln (\cos x  \sin x)+C}\)

SheldonEinstein
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oh sorry there is mod instead of brackets

amriju
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thats wrong...u should nt have played with our prestige in public...:P @hartnn

hartnn
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with a negative sign

SheldonEinstein
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why ve ?

amriju
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yeah..ofcrse

hartnn
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d/dx(cos xsin x) = sin xcos x

SheldonEinstein
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Oh ok got it :)