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

Calculus1
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multiply both sides by cos x + sin x
You will get like this : \[\large{\frac{1 + 2cosx.sinx}{cos^2x-sin^2x}}\]
Simplify n integrate.

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Other answers:

multiply num and den. by cosx + sinx...and solve for sec2x+tan2X
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.....
Very nice @hartnn. Your way is the quickest.
@hartnn so it will be : \(\large{ln (\cos x - \sin x)+C}\)
oh sorry there is mod instead of brackets
thats wrong...u should nt have played with our prestige in public...:P @hartnn
with a negative sign
why -ve ?
yeah..ofcrse
d/dx(cos x-sin x) = -sin x-cos x
Oh ok got it :)

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