sinx-cosx-sinx-cosx-dx-

Question Number 93958 by mashallah last updated on 16/May/20

\int \frac{sinx - cosx}{sinx + cosx} dx =

Commented by mathmax by abdo last updated on 17/May/20

I = \int \frac{s i n x - c o s x}{s i n x + c o s x} d x \Rightarrow I = \int \frac{t a n x - 1}{t a n x + 1} d x =_{t a n x = t}

= \int \frac{t - 1}{t + 1} \times \frac{d t}{1 + t^{2}} = \int \frac{t - 1}{(t + 1) (t^{2} + 1)} d t l e t d e c o m p o s e

F (t) = \frac{t - 1}{(t + 1) (t^{2} + 1)} \Rightarrow F (t) = \frac{a}{t + 1} + \frac{b t + c}{t^{2} + 1}

a = - 1 {l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = 1

F (o) = - 1 = a + c \Rightarrow c = 0 \Rightarrow F (t) = - \frac{1}{t + 1} + \frac{t}{t^{2} + 1} \Rightarrow

I = - l n ∣ t + 1 ∣ + \frac{1}{2} l n (1 + t^{2}) + C

= \frac{1}{2} l n (1 + {t a n}^{2} x) - l n ∣ 1 + t a n x ∣ + C

Answered by john santu last updated on 16/May/20

\frac{{(\sin x - \cos x)}^{2}}{- \cos 2 x} = \frac{1 - \sin 2 x}{- \cos 2 x}

= \tan 2 x - \sec 2 x

Answered by Kunal12588 last updated on 16/May/20

s i n x + c o s x = t

\Rightarrow - (s i n x - c o s x) d x = d t

I = \int - \frac{d t}{t} = - l n ∣ s i n x + c o s x ∣