∫^(π/2) _( 0) (((tanx))^(1/3) /((sinx+cosx)^2 ))dx


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Question Number 193439 by Nimnim111118 last updated on 14/Jun/23

${\int_{0}}^{π / 2} \frac{\sqrt[3]{t a n x}}{{(s i n x + c o s x)}^{2}} d x$
Commented by Nimnim111118 last updated on 14/Jun/23

$h e l p . . .$
Commented by Frix last updated on 14/Jun/23

$Use t = \sqrt[3]{\tan x}$ $\Rightarrow$ $3 \int \frac{t^{3}}{{(t^{3} + 1)}^{2}} d t$ $Now decompose etc$ $I get \frac{2 \sqrt{3}}{9} π$
	Answered by MM42 last updated on 14/Jun/23
	$I=∫ (((tanx))^(1/3) /((sinx+cosx)^2 ))dx=∫(((tanx))^(1/3) /(1+sin2x))dx= ∫(((tanx))^(1/3) /((1+tanx)^2 ))(1+tan^2 x)dx ; let : tanx=u^3 ⇒(1+tan^2 x)dx=3u^2 du x→0⇒u→0 & x→(π/2)⇒u→∞ ⇒I=∫((3u^3 )/((1+u^3 )^2 )) du ∫(u^3 /((1+u^3 ))) du = ((au^2 +bu+c)/(1+u^3 )) + ∫ ((a^′ u^2 +b′u+c′)/(1+u^3 )) du (ostrohrdsky) =−(2/3)×(u/(u^3 +1))+(2/3)∫ (du/(u^3 +1)) ∫ (du/(u^3 +1)) =∫ ((a/(u+1))+((bu+c)/(u^2 −u+1)))du ( decomposition of fraction) =(1/3)∫ ((1/(u+1))−((u−2)/(u^2 −u+1)))du ⇒I=−(1/3)×(u/(u^3 +1))+(2/9)×∫ ((1/(u+1))−(1/2)×((2u−1−3)/(u^2 −u+1)))du= −(1/3)×(u/(u^3 +1))+(2/9)×∫ ((1/(u+1))−(1/2)×((2u−1)/(u^2 −u+1))+(3/2)×(1/((u−(1/2))^2 +(3/4))))du= ⇒I=−(1/3)×(u/(u^3 +1))+(2/9) ln(u+1)−(1/9)ln(u^2 −u+1)+(2/(3(√3)))tan^(−1) (((2u−1)/( (√3)))) ⇒ans=−(1/3)×(u/(u^3 +1)) +(1/9)ln((((u+1)^2 )/(u^2 −u+1)))+(1/( (√3))) tan^(−1) (((2u−1)/( (√3)))) ∣_0 ^∞ =(π/( 2(√3)))+(π/( 6(√3)))=((2(√(3 ))π)/( 9 )) ✓$
	$I = \int \frac{\sqrt[3]{t a n x}}{{(s i n x + c o s x)}^{2}} d x = \int \frac{\sqrt[3]{t a n x}}{1 + s i n 2 x} d x =$ $\int \frac{\sqrt[3]{t a n x}}{{(1 + t a n x)}^{2}} (1 + {t a n}^{2} x) d x; l e t : t a n x = u^{3} \Rightarrow (1 + {t a n}^{2} x) d x = 3 u^{2} d u$ $x \to 0 \Rightarrow u \to 0 & x \to \frac{π}{2} \Rightarrow u \to \infty$ $\Rightarrow I = \int \frac{3 u^{3}}{{(1 + u^{3})}^{2}} d u$ $\int \frac{u^{3}}{(1 + u^{3})} d u = \frac{{a u}^{2} + b u + c}{1 + u^{3}} + \int \frac{a^{^{'}} u^{2} + b^{'} u + c^{'}}{1 + u^{3}} d u (o s t r o h r d s k y)$ $= - \frac{2}{3} \times \frac{u}{u^{3} + 1} + \frac{2}{3} \int \frac{d u}{u^{3} + 1}$ $\int \frac{d u}{u^{3} + 1} = \int (\frac{a}{u + 1} + \frac{b u + c}{u^{2} - u + 1}) d u (d e c o m p o s i t i o n o f f r a c t i o n)$ $= \frac{1}{3} \int (\frac{1}{u + 1} - \frac{u - 2}{u^{2} - u + 1}) d u$ $\Rightarrow I = - \frac{1}{3} \times \frac{u}{u^{3} + 1} + \frac{2}{9} \times \int (\frac{1}{u + 1} - \frac{1}{2} \times \frac{2 u - 1 - 3}{u^{2} - u + 1}) d u =$ $- \frac{1}{3} \times \frac{u}{u^{3} + 1} + \frac{2}{9} \times \int (\frac{1}{u + 1} - \frac{1}{2} \times \frac{2 u - 1}{u^{2} - u + 1} + \frac{3}{2} \times \frac{1}{{(u - \frac{1}{2})}^{2} + \frac{3}{4}}) d u =$ $\Rightarrow I = - \frac{1}{3} \times \frac{u}{u^{3} + 1} + \frac{2}{9} l n (u + 1) - \frac{1}{9} l n (u^{2} - u + 1) + \frac{2}{3 \sqrt{3}} {t a n}^{- 1} (\frac{2 u - 1}{\sqrt{3}})$ $\Rightarrow a n s = - \frac{1}{3} \times \frac{u}{u^{3} + 1} + \frac{1}{9} l n (\frac{{(u + 1)}^{2}}{u^{2} - u + 1}) + \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{2 u - 1}{\sqrt{3}}) ∣_{0}^{\infty}$ $= \frac{π}{2 \sqrt{3}} + \frac{π}{6 \sqrt{3}} = \frac{2 \sqrt{3} π}{9} ✓$
	Commented by Frix last updated on 15/Jun/23

	$I = \int \frac{3 u^{3}}{{(u^{3} + 1)}^{2}} d u$ $I = - \frac{u}{u^{3} + 1} + \frac{1}{6} \ln (\frac{{(u + 1)}^{2}}{u^{2} - u + 1}) + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 u - 1}{\sqrt{3}} + C$ $\Rightarrow$ $Your result is wrong .$
	Commented by MM42 last updated on 14/Jun/23

	$w h y ‘ ‘ I = \int \frac{3 u^{2}}{{(u^{3} + 1)}^{2}} d u^{″} ? ? ?$ $i n a d d i t i o n :$ $\int \frac{3 u^{2}}{{(u^{3} + 1)}^{2}} d u = \frac{- 1}{u^{3} + 1} + c ∵$
	Commented by MM42 last updated on 14/Jun/23

	$t h a n k y o u f o r y o u r m e n t i o n .$
	Commented by Frix last updated on 15/Jun/23

	$Typo, I corrected it 2 \to 3$
	Commented by Tawa11 last updated on 16/Jun/23

	$MM 42 you are MrW ? ?$ $I can see mrW hand writing in all MM 42 solutions .$ $Sm sorry if am wrong .$ $Just that I miss mrW$ $or you are mjs ?$
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