tan-1-sinx-1-sinx-dx-

Question Number 29043 by yesaditya22@gmail.com last updated on 03/Feb/18

\int \tan^{-} (1 - sinx / 1 + sinx) dx

Commented by abdo imad last updated on 03/Feb/18

l e t p u t I = \int a r c t a n (\frac{1 - s i n x}{1 + s i n x}) d x (a r c t a n = {t a n}^{- 1}) w e h a v e

\frac{1 - s i n x}{1 + s i n x} = \frac{1 - c o s (\frac{π}{2} - x)}{1 + c o s (\frac{π}{2} - x)} = \frac{2 {s i n}^{2} (\frac{π}{4} - \frac{x}{2})}{2 {c o s}^{2} (\frac{π}{4} - \frac{x}{2})} = {t a n}^{2} (\frac{π}{4} - \frac{x}{2})

I = \int a r c t a n ({t a n}^{2} (\frac{π}{4} - \frac{x}{2})) d x l e t u s e t h e c h .

{t a n}^{2} (\frac{π}{4} - \frac{x}{2}) = u \Leftrightarrow t a n (\frac{π}{4} - \frac{x}{2}) = \sqrt{u} a n d

\frac{π}{4} - \frac{x}{2} = a r c t a n (\sqrt{u}) \Leftrightarrow x = - 2 a r c t a n (\sqrt{u}) \Rightarrow

d x = - 2 \frac{\frac{1}{2 \sqrt{u}}}{1 + u} = \frac{- 1}{\sqrt{u} (1 + u)} s o

I = - \int \frac{a r t a n u}{\sqrt{u} (1 + u)} d u (\sqrt{u} = t)

I = - \int \frac{a r c t a n (t^{2})}{t (1 + t^{2})} 2 t d t = - 2 \int \frac{a r c t a n (t^{2})}{1 + t^{2}} d t + λ a n d t h e

i n t e g r a l \int \frac{a r c t a n (t^{2})}{1 + t^{2}} d t b e c a l c u l a t e d b y t h e p a r a m e t r i c

f u n c t i o n f (x) = \int \frac{a r c t a n (x (1 + t^{2}))}{1 + t^{2}} d t b y d e r i v a t i o n u n d e r

\int \dots .