I = ∫_0 ^(π/2) tan^(−1) (((sinx )/2))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 199570 by universe last updated on 05/Nov/23

$I = \int_{0}^{π / 2} \tan^{- 1} (\frac{\sin x}{2}) d x$
	Answered by witcher3 last updated on 05/Nov/23
	$I(a)=∫_0 ^(π/2) tan^(−1) (asin(x))dx I=I((1/2)) I(0)=0 I′(a)=∫_0 ^(π/2) ((sin(x))/(1+a^2 sin^2 (x)))dx =∫_0 ^1 (dy/(1+a^2 (1−y^2 )))dy =∫_0 ^1 (dy/( [(√(1+a^2 ))−ay][(√(a^2 +1))+ay])) =(1/(2(√(a^2 +1))))∫_0 ^1 {(1/( (√(1+a^2 ))−ay))+(1/(ay+(√(1+a^2 ))))}dy =(1/(2a(√(1+a^2 ))))ln(((a+(√(1+a^2 )))/( (√(1+a^2 ))−a)))=I′(a) I((1/2))=∫_0 ^(1/2) (1/(2a(√(1+a^2 ))))ln(((a+(√(1+a^2 )))/( (√(1+a^2 ))−a)))da a=sh(x)⇒da=ch(x)dx I=∫_0 ^(sh^− ((1/2))) (1/(2sh(x)))ln((e^x /e^(−x) ))=∫_0 ^(sh^− ((1/2))) (x/(e^x −e^(−x) ))dx =(1/2)∫_0 ^(sh^− ((1/2))) (x/(e^x −1))+(x/(e^x +1))dx=I ∫(x/(e^x −1))=xln(1−e^(−x) )−∫((ln(1−e^(−x) )de^(−x) )/e^(−x) ) =xln(1−e^(−x) )−Li_2 (e^(−x) ) ∫(x/(e^x +1))dx=−xln(e^(−x) +1)+Li_2 (−e^(−x) ) I=(1/2)(sh^− ((1/2))ln(1−e^(−sh^− ((1/2))) )−Li_2 (e^(−sh^− ((1/2))) ) −sh^− ((1/2))ln(1+e^(−sh^− ((1/2))) )+Li_2 (−e^(−sh^− ((1/2))) )+Li_2 (1)−Li_2 (−1) =(1/(12))(π^2 −6sinh^− (2)csch^− (2))$
	$I (a) = \int_{0}^{\frac{π}{2}} \tan^{- 1} (asin (x)) dx$ $I = I (\frac{1}{2})$ $I (0) = 0$ $I^{'} (a) = \int_{0}^{\frac{π}{2}} \frac{\sin (x)}{1 + a^{2} \sin^{2} (x)} dx$ $= \int_{0}^{1} \frac{dy}{1 + a^{2} (1 - y^{2})} dy$ $= \int_{0}^{1} \frac{dy}{[\sqrt{1 + a^{2}} - ay] [\sqrt{a^{2} + 1} + ay]}$ $= \frac{1}{2 \sqrt{a^{2} + 1}} \int_{0}^{1} {\frac{1}{\sqrt{1 + a^{2}} - ay} + \frac{1}{ay + \sqrt{1 + a^{2}}}} dy$ $= \frac{1}{2 a \sqrt{1 + a^{2}}} \ln (\frac{a + \sqrt{1 + a^{2}}}{\sqrt{1 + a^{2}} - a}) = I^{'} (a)$ $I (\frac{1}{2}) = \int_{0}^{\frac{1}{2}} \frac{1}{2 a \sqrt{1 + a^{2}}} \ln (\frac{a + \sqrt{1 + a^{2}}}{\sqrt{1 + a^{2}} - a}) da$ $a = sh (x) \Rightarrow da = ch (x) dx$ $I = \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{1}{2 sh (x)} \ln (\frac{e^{x}}{e^{- x}}) = \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{x}{e^{x} - e^{- x}} dx$ $= \frac{1}{2} \int_{0}^{{sh}^{-} (\frac{1}{2})} \frac{x}{e^{x} - 1} + \frac{x}{e^{x} + 1} dx = I$ $\int \frac{x}{e^{x} - 1} = xln (1 - e^{- x}) - \int \frac{\ln (1 - e^{- x}) {de}^{- x}}{e^{- x}}$ $= xln (1 - e^{- x}) - {Li}_{2} (e^{- x})$ $\int \frac{x}{e^{x} + 1} dx = - xln (e^{- x} + 1) + {Li}_{2} (- e^{- x})$ $I = \frac{1}{2} ({sh}^{-} (\frac{1}{2}) \ln (1 - e^{- {sh}^{-} (\frac{1}{2})}) - {Li}_{2} (e^{- {sh}^{-} (\frac{1}{2})})$ $- {sh}^{-} (\frac{1}{2}) \ln (1 + e^{- {sh}^{-} (\frac{1}{2})}) + {Li}_{2} (- e^{- {sh}^{-} (\frac{1}{2})}) + {Li}_{2} (1) - L i_{2} (- 1)$ $= \frac{1}{12} (π^{2} - {6 \sinh}^{-} (2) {csch}^{-} (2))$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum