∫^( 1/64) _0 ((tan^(−1) x)/( (√x)))dx


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Question Number 131654 by mohammad17 last updated on 07/Feb/21

${\int_{0}}^{1 / 64} \frac{{t a n}^{- 1} x}{\sqrt{x}} d x$
	Answered by mathmax by abdo last updated on 07/Feb/21
	$I=∫_0 ^(1/(64)) ((arctanx)/( (√x)))dx ⇒I =_((√x)=t) ∫_0 ^(1/8) ((arctan(t^2 ))/t)(2t)dt =2∫_0 ^(1/8) arctan(t^2 )dt =2{[tarctan(t^2 )]_0 ^(1/8) −∫_0 ^(1/8) t.((2t)/(1+t^4 ))} =2((1/8)arcctan((1/(64)))−2∫_0 ^(1/8) (t^2 /(1+t^4 ))dt)=(1/4)arctan((1/(64)))−4∫_0 ^(1/8) (t^2 /(1+t^4 ))dt we have ∫ (t^2 /(1+t^4 ))dt =∫ (1/((1/t^2 )+t^2 )) dt=(1/2) ∫ ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =(1/2)∫ ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt(→t+(1/t)=u)+(1/2)∫((1+(1/t^2 ))/((t−(1/t))^2 +2))(t−(1/t)=v) =(1/2)∫ (du/(u^2 −2)) +(1/2)∫ (dv/(v^2 +2))(v=(√2)y) =(1/(4(√2)))∫ ((1/(u−(√2)))−(1/(u+(√2))))du +(1/2)∫(((√2)dy)/(2(1+y^2 ))) =(1/(4(√2)))ln∣((u−(√2))/(u+(√2)))∣ +(1/(2(√2)))arctany +c =(1/(4(√2)))ln∣((t+(1/t)−(√2))/(t+(1/t)+(√2)))∣+(1/(2(√2)))arctan((1/( (√2)))(t−(1/t)))⇒ ∫_0 ^(1/8) (t^2 /(1+t^4 ))dt =(1/(4(√2)))[ln(((t^2 −(√2)t+1)/(t^2 +(√2)t+1)))]_0 ^(1/8) +(1/(2(√2)))[arctan(((t^2 −1)/(t(√2))))]_0 ^(1/8) =(1/(4(√2)))ln((((1/(64))−((√2)/8)+1)/((1/(64))+((√2)/( 8))+1)))+(1/(2(√2)))(arctan((((1/(64))−1)/( (√2)))×8 +(π/2)))=...$
	$I = \int_{0}^{\frac{1}{64}} \frac{arctanx}{\sqrt{x}} dx \Rightarrow I =_{\sqrt{x} = t} \int_{0}^{\frac{1}{8}} \frac{\arctan (t^{2})}{t} (2 t) dt$ $= 2 \int_{0}^{\frac{1}{8}} \arctan (t^{2}) dt = 2 {{[tarctan (t^{2})]}_{0}^{\frac{1}{8}} - \int_{0}^{\frac{1}{8}} t . \frac{2 t}{1 + t^{4}}}$ $= 2 (\frac{1}{8} arcctan (\frac{1}{64}) - 2 \int_{0}^{\frac{1}{8}} \frac{t^{2}}{1 + t^{4}} dt) = \frac{1}{4} \arctan (\frac{1}{64}) - 4 \int_{0}^{\frac{1}{8}} \frac{t^{2}}{1 + t^{4}} dt$ $we have \int \frac{t^{2}}{1 + t^{4}} dt = \int \frac{1}{\frac{1}{t^{2}} + t^{2}} dt = \frac{1}{2} \int \frac{1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt$ $= \frac{1}{2} \int \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt (\to t + \frac{1}{t} = u) + \frac{1}{2} \int \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} (t - \frac{1}{t} = v)$ $= \frac{1}{2} \int \frac{du}{u^{2} - 2} + \frac{1}{2} \int \frac{dv}{v^{2} + 2} (v = \sqrt{2} y)$ $= \frac{1}{4 \sqrt{2}} \int (\frac{1}{u - \sqrt{2}} - \frac{1}{u + \sqrt{2}}) du + \frac{1}{2} \int \frac{\sqrt{2} dy}{2 (1 + y^{2})}$ $= \frac{1}{4 \sqrt{2}} \ln ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣ + \frac{1}{2 \sqrt{2}} arctany + c$ $= \frac{1}{4 \sqrt{2}} \ln ∣ \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} ∣ + \frac{1}{2 \sqrt{2}} \arctan (\frac{1}{\sqrt{2}} (t - \frac{1}{t})) \Rightarrow$ $\int_{0}^{\frac{1}{8}} \frac{t^{2}}{1 + t^{4}} dt = \frac{1}{4 \sqrt{2}} {[\ln (\frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1})]}_{0}^{\frac{1}{8}} + \frac{1}{2 \sqrt{2}} {[\arctan (\frac{t^{2} - 1}{t \sqrt{2}})]}_{0}^{\frac{1}{8}}$ $= \frac{1}{4 \sqrt{2}} \ln (\frac{\frac{1}{64} - \frac{\sqrt{2}}{8} + 1}{\frac{1}{64} + \frac{\sqrt{2}}{8} + 1}) + \frac{1}{2 \sqrt{2}} (\arctan (\frac{\frac{1}{64} - 1}{\sqrt{2}} \times 8 + \frac{π}{2})) = . . .$
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