klimenkov
  • klimenkov
What about this? \[\int{\frac{\sin x+\cos x}{\cos x - \sin x}}dx\]
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
multiply both sides by cos x + sin x
anonymous
  • anonymous
You will get like this : \[\large{\frac{1 + 2cosx.sinx}{cos^2x-sin^2x}}\]
anonymous
  • anonymous
Simplify n integrate.

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

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