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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
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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SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
multiply both sides by cos x + sin x
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
You will get like this : \[\large{\frac{1 + 2cosx.sinx}{cos^2xsin^2x}}\]
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
Simplify n integrate.
 one year ago

amriju Group TitleBest ResponseYou've already chosen the best response.0
multiply num and den. by cosx + sinx...and solve for sec2x+tan2X
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.4
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.....
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Very nice @hartnn. Your way is the quickest.
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
@hartnn so it will be : \(\large{ln (\cos x  \sin x)+C}\)
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
oh sorry there is mod instead of brackets
 one year ago

amriju Group TitleBest ResponseYou've already chosen the best response.0
thats wrong...u should nt have played with our prestige in public...:P @hartnn
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.4
with a negative sign
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
why ve ?
 one year ago

amriju Group TitleBest ResponseYou've already chosen the best response.0
yeah..ofcrse
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.4
d/dx(cos xsin x) = sin xcos x
 one year ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.0
Oh ok got it :)
 one year ago
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