Question

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

Answer 1

multiply both sides by cos x + sin x

Answer 2

You will get like this :

\[\large{\frac{1 + 2cosx.sinx}{cos^2x-sin^2x}}\]

Answer 3

Simplify n integrate.

Answer 4

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

Answer 5

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.....

Answer 6

Very nice @hartnn. Your way is the quickest.

Answer 7

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

Answer 8

oh sorry there is mod instead of brackets

Answer 9

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

Answer 10

with a negative sign

Answer 11

why -ve ?

Answer 12

yeah..ofcrse

Answer 13

d/dx(cos x-sin x) = -sin x-cos x

Answer 14

Oh ok got it :)