Question Number 131654 by mohammad17 last updated on 07/Feb/21
$$\underset{\mathrm{0}} {\int}^{\:\mathrm{1}/\mathrm{64}} \frac{{tan}^{−\mathrm{1}} {x}}{\:\sqrt{{x}}}{dx} \\ $$
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)))=...](Q131670.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{64}}} \:\frac{\mathrm{arctanx}}{\:\sqrt{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:=\mathrm{2}\left\{\left[\mathrm{tarctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} −\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \mathrm{t}.\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\right\} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arcctan}\left(\frac{\mathrm{1}}{\mathrm{64}}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{64}}\right)−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{have}\:\int\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\int\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{2}}\mathrm{dt}\left(\rightarrow\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{u}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{v}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} −\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} \:+\mathrm{2}}\left(\mathrm{v}=\sqrt{\mathrm{2}}\mathrm{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int\:\left(\frac{\mathrm{1}}{\mathrm{u}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{u}+\sqrt{\mathrm{2}}}\right)\mathrm{du}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{\mathrm{2}}\mathrm{dy}}{\mathrm{2}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{u}−\sqrt{\mathrm{2}}}{\mathrm{u}+\sqrt{\mathrm{2}}}\mid\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{arctany}\:+\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}−\sqrt{\mathrm{2}}}{\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}+\sqrt{\mathrm{2}}}\mid+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\left(\frac{\mathrm{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{t}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{t}+\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{arctan}\left(\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{t}\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{8}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\frac{\mathrm{1}}{\mathrm{64}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}+\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{64}}+\frac{\sqrt{\mathrm{2}}}{\:\mathrm{8}}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{arctan}\left(\frac{\frac{\mathrm{1}}{\mathrm{64}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\mathrm{8}\:+\frac{\pi}{\mathrm{2}}\right)\right)=... \\ $$