advanced-calculus-evaluate-I-0-pi-2-ln-tan-x-sin-x-cos-x-2-dx-m-n-

Question Number 121869 by mnjuly1970 last updated on 12/Nov/20

\dots a d v a n c e d c a l c u l u s \dots

e v a l u a t e ::

I \overset{? ? ?}{=} \int_{0}^{\frac{π}{2}} {(\frac{l n (t a n (x))}{s i n (x) - c o s (x)})}^{2} d x

. m . n .

Answered by mindispower last updated on 12/Nov/20

p r o b l e m i n π / 4 o r P v . . ?

Commented by mnjuly1970 last updated on 13/Nov/20

0 t o \frac{π}{2} i s c o r r e c t \dots

Answered by mindispower last updated on 13/Nov/20

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

I = \int_{0}^{\frac{π}{2}} \frac{{l n}^{2} (t g (x))}{{(t g (x) - 1)}^{2}} d t g (x)

t g (x) = t \Rightarrow I = \int_{0}^{\infty} \frac{{l n}^{2} (r)}{{(r - 1)}^{2}} d r

= \int_{0}^{1} (\frac{{l n}^{2} (r)}{{(r - 1)}^{2}} . d r + \frac{{l n}^{2} (\frac{1}{r})}{{(\frac{1}{r} - 1)}^{2}} . \frac{d r}{r^{2}})

\int_{0}^{1} \frac{{l n}^{2} (r) d r}{{(r - 1)}^{2}} . . I B P \Rightarrow

\lim_{t \to 0} {[\frac{- {l n}^{2} (r)}{r - 1}]}_{t}^{1} + 2 \int_{t}^{1} \frac{l n (r)}{r (r - 1)} d r

= \lim_{t \to 0} \frac{{l n}^{2} (t)}{t - 1} + 2 \int_{t}^{1} \frac{l n (r)}{r - 1} d r - 2 \int_{t}^{1} \frac{l n (r)}{r} d r

= \lim_{t \to 0} \frac{{l n}^{2} (t)}{t - 1} - 2 \int_{0}^{1 - t} \frac{l n (1 - r)}{r} d r + {l n}^{2} (t)

= \lim_{t \to 0} [(\frac{{l n}^{2} (t)}{t - 1} + {l n}^{2} (t)) + 2 {L i}_{2} (1 - t)]

= \lim_{t \to 0} [\frac{{t l n}^{2} (t)}{t - 1}] + 2 {L i}_{2} (1) c a u s e {L i}_{2} i s c o n t i n u s

= 2 {L i}_{2} (1) = 2 . \frac{π^{2}}{6} = \frac{π^{2}}{3}

Commented by mnjuly1970 last updated on 13/Nov/20

b r a v o b r a v o

s i r m i n d s p o w e r

t h a n k y o u s o m u c h \dots

Commented by mindispower last updated on 13/Nov/20

t h a n x a l w a y s a p l e a s u r