...nice calculus... prove that:: ∫_((−π)/4) ^(π/4) (((π−4x)tan(x))/(1−tan(x)))dx=^(???) πln(2)−(π^2 /4)


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

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $\int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x \overset{? ? ?}{=} π l n (2) - \frac{π^{2}}{4}$
	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

	$t h a n k y o u s o m u c h . . .$
	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
	$solution:I=∫_((−π)/4) ^( (π/4)) (((π−4x)tan(x))/(1−tan(x)))dx note :: ∫_0 ^(π/2) ln(sin(x))dx=((−π)/2)ln(2) =^(u=(π/4)−x) 4∫_0 ^( (π/2)) ((xtan((π/4)−u))/(1−tan((π/4)−u)))du =4∫_0 ^( (π/2)) {((x((1−tan(u))/(1+tan(u))))/((2tan(x))/(1+tan(x))))}du =2∫_0 ^( (π/2)) xcot(x)(1−tan(x))dx =2∫_0 ^( (π/2)) xcot(x)dx−((π^2 /4)) =2{[xln(sin(x))]_0 ^(π/2) −∫_0 ^( (π/2)) ln(sin(x))dx}−(π^2 /4) =−2∫_0 ^( (π/2)) ln(sin(x))dx−(π^2 /4) =−2(((−π)/2)ln(2))−(π^2 /4)=πln(2)−(π^2 /4) ∴ ∫_((−π )/4) ^( (π/4)) (((π−4x)tan(x))/(1−tan(x)))dx=πln(2)−(π^2 /4) ✓$
	$s o l u t i o n : I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x$ $n o t e :: \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x = \frac{- π}{2} l n (2)$ $\overset{u = \frac{π}{4} - x}{=} 4 \int_{0}^{\frac{π}{2}} \frac{x t a n (\frac{π}{4} - u)}{1 - t a n (\frac{π}{4} - u)} d u$ $= 4 \int_{0}^{\frac{π}{2}} {\frac{x \frac{1 - t a n (u)}{1 + t a n (u)}}{\frac{2 t a n (x)}{1 + t a n (x)}}} d u$ $= 2 \int_{0}^{\frac{π}{2}} x c o t (x) (1 - t a n (x)) d x$ $= 2 \int_{0}^{\frac{π}{2}} x c o t (x) d x - (\frac{π^{2}}{4})$ $= 2 {{[x l n (s i n (x))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x} - \frac{π^{2}}{4}$ $= - 2 \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x - \frac{π^{2}}{4}$ $= - 2 (\frac{- π}{2} l n (2)) - \frac{π^{2}}{4} = π l n (2) - \frac{π^{2}}{4}$ $∴ \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{(π - 4 x) t a n (x)}{1 - t a n (x)} d x = π l n (2) - \frac{π^{2}}{4} ✓$
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