how-do-we-caculate-ln-tanx-dx-

Question Number 133734 by Abdoulaye last updated on 23/Feb/21

h o w d o w e c a c u l a t e \int l n (t a n x) d x ?

Answered by mathmax by abdo last updated on 24/Feb/21

f (x) = \int_{\frac{π}{4}}^{x} \ln (tanx) dx \Rightarrow f (x) =_{tanx = t} \int_{1}^{tanx} \frac{\ln (t)}{1 + t^{2}} dt

= {[\arctan (t) \ln (t)]}_{1}^{tanx} - \int_{1}^{tanx} \frac{arctant}{t} dt

= xln (tanx) - \int_{1}^{tanx} \frac{\arctan (t)}{t} dt we have {(arctant)}^{(1)} = \frac{1}{1 + t^{2}}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{2 n} \Rightarrow \arctan (t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n + 1} + c (c = 0) \Rightarrow

\frac{arctant}{t} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n} \Rightarrow f (x) = xln (x) - \int_{1}^{tanx} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n} dt

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} {[\frac{1}{2 n + 1} t^{2 n + 1}]}_{1}^{tanx}

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} (\tan^{2 n + 1} x - 1)

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} \tan^{2 n + 1} x - K (katalan constant) \dots . be continued \dots

Commented by Abdoulaye last updated on 24/Feb/21

t h a n k

p l e a s e w h a t k a t a l a n c a n s t a n t ?

Commented by Dwaipayan Shikari last updated on 24/Feb/21

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = 1 - \frac{1}{3^{2}} + \frac{1}{5^{2}} - \frac{1}{7^{2}} + \dots . = G ({C a t a l a n}^{'} s c o n s t a n t)