nice-calculus-prove-that-pi-4-pi-4-pi-4x-tan-x-1-tan-x-dx-piln-2-pi-2-4-

Question Number 123937 by mnjuly1970 last updated on 29/Nov/20

\dots n i c e c a l c u l u s \dots

p r o v e t h a t ::

\int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x \overset{? ? ?}{=} π l n (2) - \frac{π^{2}}{4}

Answered by Dwaipayan Shikari last updated on 29/Nov/20

\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n x}{1 - t a n x} d x = 4 \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x (1 - t a n x)}{2 t a n x}

= 4 \int_{0}^{\frac{π}{4}} x c o t x - x d x

= 2 \int_{0}^{\frac{π}{2}} x c o t x - x d x = 2 x {[l o g (s i n x)]}_{0}^{\frac{π}{2}} - 2 \int_{0}^{\frac{π}{2}} l o g (s i n x) - \frac{π^{2}}{4}

= - 2 . (- \frac{π}{2} l o g (2)) - \frac{π^{2}}{4} = π l o g (2) - \frac{π^{2}}{4} = - 0.289815

Commented by mnjuly1970 last updated on 29/Nov/20

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

Commented by CanovasCamiseros last updated on 29/Nov/20

P l e a s e w h o c a n h e l p m e w i t h h a r d d i f f e r e n t i a l e q u a t i o n s q u e s t i o n s ?

Commented by Dwaipayan Shikari last updated on 29/Nov/20

K i n d l y p o s t y o u r p r o b l e m s i r!

Answered by mnjuly1970 last updated on 29/Nov/20

s o l u t i o n : I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x

n o t e :: \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x = \frac{- π}{2} l n (2)

\overset{u = \frac{π}{4} - x}{=} 4 \int_{0}^{\frac{π}{2}} \frac{x t a n (\frac{π}{4} - u)}{1 - t a n (\frac{π}{4} - u)} d u

= 4 \int_{0}^{\frac{π}{2}} {\frac{x \frac{1 - t a n (u)}{1 + t a n (u)}}{\frac{2 t a n (x)}{1 + t a n (x)}}} d u

= 2 \int_{0}^{\frac{π}{2}} x c o t (x) (1 - t a n (x)) d x

= 2 \int_{0}^{\frac{π}{2}} x c o t (x) d x - (\frac{π^{2}}{4})

= 2 {{[x l n (s i n (x))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x} - \frac{π^{2}}{4}

= - 2 \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x - \frac{π^{2}}{4}

= - 2 (\frac{- π}{2} l n (2)) - \frac{π^{2}}{4} = π l n (2) - \frac{π^{2}}{4}

∴ \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x = π l n (2) - \frac{π^{2}}{4} ✓