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Question Number 113821 by 675480065 last updated on 15/Sep/20
∫_0 ^(π/2) ln(2−sinx)dx
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}−\mathrm{sinx}\right)\mathrm{dx} \\ $$
Commented by Dwaipayan Shikari last updated on 15/Sep/20
I(a)=∫_0 ^(π/2) log(2+asinx)dx I′(a)=∫_0 ^(π/2) ((sinx)/(2+asinx))dx I′(a)=(1/a)∫_0 ^(π/2) 1−(2/(2+asinx)) I′(a)=(π/(2a))−2∫_0 ^(π/2) (1/(2+asinx))dx I′(a)=(π/(2a))−4∫_0 ^(π/2) (1/(2+((2at)/(1+t^2 )))).(1/(1+t^2 ))dt (tan(x/2)=t) I′(a)=(π/(2a))−4∫_0 ^1 (1/(2+2t^2 +2at))dt I′(a)=(π/(2a))−2∫_0 ^1 (1/((t+(a/2))^2 +1−(a^2 /4)))dt I′(a)=(π/(2a))−2(1/( (√(1−(a^2 /4))))).[tan^(−1) ((2t+a)/( (√(4−a^2 ))))]_0 ^1 I(a)=(π/2)log(a)−∫(4/( (√(4−a^2 ))))(tan^(−1) (√((2+a)/(2−a))) −tan^(−1) (a/( (√(4−a^2 ))))).... .....
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{2}+{asinx}\right){dx} \\ $$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}}{\mathrm{2}+{asinx}}{dx} \\ $$$${I}'\left({a}\right)=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}+{asinx}} \\ $$$${I}'\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}+{asinx}}{dx} \\ $$$${I}'\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{2}{at}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({tan}\frac{{x}}{\mathrm{2}}={t}\right) \\ $$$${I}'\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{at}}{dt} \\ $$$${I}'\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({t}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}{dt} \\ $$$${I}'\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}−\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}}.\left[{tan}^{−\mathrm{1}} \frac{\mathrm{2}{t}+{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${I}\left({a}\right)=\frac{\pi}{\mathrm{2}}{log}\left({a}\right)−\int\frac{\mathrm{4}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\left({tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}+{a}}{\mathrm{2}−{a}}}\:−{tan}^{−\mathrm{1}} \frac{{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right)…. \\ $$$$….. \\ $$
Answered by mathdave last updated on 15/Sep/20
solution let A=∫_0 ^(π/2) ln[2(1−(1/2)sinx)]dx=ln2∫^(π/2) dx+∫^(π/2) (1−(1/2)sinx)dx A=(π/2)ln2+∫_0 ^(π/2) (1−(1/2)sinx)dx let I=∫_0 ^(π/2) (1−(1/2)sinx)dx.........(1)
$${solution}\: \\ $$$${let} \\ $$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left[\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right)\right]{dx}=\mathrm{ln2}\int^{\frac{\pi}{\mathrm{2}}} {dx}+\int^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx} \\ $$$${A}=\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx} \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}………\left(\mathrm{1}\right) \\ $$
Commented by mathdave last updated on 15/Sep/20
Commented by mathdave last updated on 15/Sep/20
Commented by mathdave last updated on 15/Sep/20
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by Olaf last updated on 15/Sep/20
I = ∫_0 ^(π/2) ln(2−sinx)dx (1/(1−(1/2)sinx)) = Σ_(n=0) ^∞ ((sin^n x)/2^n ) ((−(1/2)cosx)/(1−(1/2)sinx)) = −(1/2)Σ_(n=0) ^∞ cosx((sin^n x)/2^n ) ln(1−(1/2)sinx) = −(1/2)Σ_(n=0) ^∞ ((sin^(n+1) x)/(2^n (n+1))) I = (π/2)ln2+∫_0 ^(π/2) ln(1−(1/2)sinx)dx I = (π/2)ln2−∫_0 ^(π/2) Σ_(n=0) ^∞ ((sin^(n+1) x)/(2^(n+1) (n+1)))dx I = (π/2)ln2−∫_0 ^(π/2) Σ_(n=1) ^∞ ((sin^n x)/(n2^n ))dx I = (π/2)ln2−Σ_(n=1) ^∞ (1/(n2^n ))∫_0 ^(π/2) sin^n xdx W_n = ∫_0 ^(π/2) sin^n xdx (Wallis) I = (π/2)ln2−Σ_(n=1) ^∞ (W_n /(n2^n ))...
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}−\mathrm{sin}{x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}^{{n}} {x}}{\mathrm{2}^{{n}} } \\ $$$$\frac{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{cos}{x}\frac{\mathrm{sin}^{{n}} {x}}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{\mathrm{2}^{{n}} \left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{\mathrm{2}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)}{dx} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}^{{n}} {x}}{{n}\mathrm{2}^{{n}} }{dx} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {xdx} \\ $$$$\mathrm{W}_{{n}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {xdx}\:\left(\mathrm{Wallis}\right) \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{W}_{{n}} }{{n}\mathrm{2}^{{n}} }… \\ $$
Answered by mathmax by abdo last updated on 15/Sep/20
let I =∫_0 ^(π/2) ln(2−sinx)dx ⇒I =∫_0 ^(π/2) ln(2)+ln(1−(1/2)sinx)dx =(π/2)ln(2) +∫_0 ^(π/2) ln(1−(1/2)sinx)dx let f(a) =∫_0 ^(π/2) ln(1−asinx)dx with0<a<1 f^′ (a) =∫_0 ^(π/2) ((−sinx)/(1−asinx)) dx =_(tan((x/2))=t) −∫_0 ^1 (((2t)/(1+t^2 ))/(1−a((2t)/(1+t^2 ))))×((2dt)/(1+t^2 )) =−∫_0 ^1 ((4t)/((1+t^2 )(1+t^2 −2at))) dt =−4 ∫_0 ^1 ((tdt)/((t^2 +1)(t^2 −2at +1))) let decomoose F(t) =(t/((t^2 −2at+1)(t^2 +1))) t^2 −2at +1=0→Δ^′ =a^2 −1<0 ⇒F(t) =((αt +β)/(t^2 −2at +1)) +((mt +n)/(t^(2 ) +1)) ⇒(αt+β)(t^2 +1)+(mt+n)(t^2 −2at +1) =t ⇒ αt^3 +αt +βt^2 +β +mt^3 −2amt^2 +mt +nt^2 −2ant +n =t ⇒ (α+m)t^3 +(β−2am +n)t^2 +(α+m−2an)t +β+n =t ⇒ { ((α+m =0)),((β−2am +n =0 and { ((α+m −2an =1)),((β+n =0 ⇒)) :})) :} { ((m=−α and { ((−2an =1)),((β =−n ⇒ n =−(1/(2a)))) :})),((−2aα =0 )) :} and α=0 ⇒m=0 ⇒F(t) =((1/(2a))/(t^2 −2at +1)) +((−(1/(2a)))/(t^2 +1)) =(1/(2a)){(1/(t^2 −2at +1))−(1/(t^2 +1))} ⇒ f^′ (a) =−(2/a)∫_0 ^1 (dt/(t^2 −2at +1)) +(2/a) ∫_0 ^1 (dt/(t^2 +1)) =(2/a)×(π/4) −(2/a) ∫_0 ^1 (dt/(t^2 −2at +1)) =(π/(2a))−(2/a) ∫_0 ^1 (dt/(t^2 −2at +1)) but ∫_0 ^1 (dt/(t^2 −2at +1)) =∫_0 ^1 (dt/(t^2 −2at +a^2 +1−a^2 )) =∫_0 ^1 (dt/((t−a)^2 +1−a^2 )) =_(t−a =(√(1−a^2 ))u) ∫_((−a)/( (√(1−a^2 )))) ^((1−a)/( (√(1−a^2 )))) ((√(1−a^2 ))/((1−a^2 )(1+u^2 )))du =(1/( (√(1−a^2 )))){ arctan(((1−a)/( (√(1−a^2 )))))+arctan((a/( (√(1−a^2 )))))} ⇒ f(a) =(π/2)lna −2 ∫ (1/(a(√(1−a^2 )))) arctan(((1−a)/( (√(1−a^2 )))))da−2∫(1/(a(√(1−a^2 )))) arctan((a/( (√(1−a^2 )))))da ....be continued ....
$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}−\mathrm{sinx}\right)\mathrm{dx}\:\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinx}\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinx}\right)\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\mathrm{asinx}\right)\mathrm{dx} \\ $$$$\mathrm{with0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−\mathrm{sinx}}{\mathrm{1}−\mathrm{asinx}}\:\mathrm{dx}\:=_{\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{a}\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}×\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{4t}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\right)}\:\mathrm{dt}\:\:=−\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{tdt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}\right)} \\ $$$$\mathrm{let}\:\mathrm{decomoose}\:\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{t}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2at}+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} \:=\mathrm{a}^{\mathrm{2}} −\mathrm{1}<\mathrm{0}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\alpha\mathrm{t}\:+\beta}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}\:+\frac{\mathrm{mt}\:+\mathrm{n}}{\mathrm{t}^{\mathrm{2}\:} \:+\mathrm{1}} \\ $$$$\Rightarrow\left(\alpha\mathrm{t}+\beta\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)+\left(\mathrm{mt}+\mathrm{n}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}\right)\:=\mathrm{t}\:\Rightarrow \\ $$$$\alpha\mathrm{t}^{\mathrm{3}} \:+\alpha\mathrm{t}\:+\beta\mathrm{t}^{\mathrm{2}} \:+\beta\:\:+\mathrm{mt}^{\mathrm{3}} −\mathrm{2amt}^{\mathrm{2}} +\mathrm{mt}\:+\mathrm{nt}^{\mathrm{2}} −\mathrm{2ant}\:+\mathrm{n}\:=\mathrm{t}\:\Rightarrow \\ $$$$\left(\alpha+\mathrm{m}\right)\mathrm{t}^{\mathrm{3}} \:+\left(\beta−\mathrm{2am}\:+\mathrm{n}\right)\mathrm{t}^{\mathrm{2}} \:+\left(\alpha+\mathrm{m}−\mathrm{2an}\right)\mathrm{t}\:+\beta+\mathrm{n}\:=\mathrm{t}\:\Rightarrow \\ $$$$\begin{cases}{\alpha+\mathrm{m}\:=\mathrm{0}}\\{\beta−\mathrm{2am}\:+\mathrm{n}\:=\mathrm{0}\:\:\:\:\mathrm{and}\:\begin{cases}{\alpha+\mathrm{m}\:−\mathrm{2an}\:=\mathrm{1}}\\{\beta+\mathrm{n}\:=\mathrm{0}\:\:\Rightarrow}\end{cases}}\end{cases} \\ $$$$\begin{cases}{\mathrm{m}=−\alpha\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\begin{cases}{−\mathrm{2an}\:=\mathrm{1}}\\{\beta\:=−\mathrm{n}\:\:\:\:\Rightarrow\:\mathrm{n}\:=−\frac{\mathrm{1}}{\mathrm{2a}}}\end{cases}}\\{−\mathrm{2a}\alpha\:=\mathrm{0}\:\:}\end{cases} \\ $$$$\mathrm{and}\:\alpha=\mathrm{0}\:\Rightarrow\mathrm{m}=\mathrm{0}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{2a}}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}\:+\frac{−\frac{\mathrm{1}}{\mathrm{2a}}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2a}}\left\{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=−\frac{\mathrm{2}}{\mathrm{a}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}\:+\frac{\mathrm{2}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}}×\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{2}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}\:=\frac{\pi}{\mathrm{2a}}−\frac{\mathrm{2}}{\mathrm{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}} \\ $$$$\mathrm{but}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2at}\:+\mathrm{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\left(\mathrm{t}−\mathrm{a}\right)^{\mathrm{2}} \:+\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$=_{\mathrm{t}−\mathrm{a}\:=\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\mathrm{u}} \:\:\:\int_{\frac{−\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}−\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}} \:\:\:\:\:\:\frac{\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\left\{\:\mathrm{arctan}\left(\frac{\mathrm{1}−\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)+\mathrm{arctan}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}\mathrm{lna}\:\:−\mathrm{2}\:\int\:\:\frac{\mathrm{1}}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{1}−\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\mathrm{da}−\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\mathrm{da} \\ $$$$….\mathrm{be}\:\mathrm{continued}\:…. \\ $$
Commented by mathmax by abdo last updated on 16/Sep/20
we have ∫ (1/(a(√(1−a^2 )))) arctan(((1−a)/( (√(1−a^2 )))))da =_(a =cost) ∫ (1/(cost sint)) arctan(((1−cost)/(sint)))(−sint)dt =−∫ (1/(cost)) arctan(((2sin^2 ((t/2)))/(2sin((t/2))cos((t/2)))))dt =−∫ (1/(cost)) arctan(tan((t/2))dt =−(1/2) ∫ (t/(cost)) dt =_(tan((t/2))=z) −(1/2) ∫ ((2arctan(z))/((1−z^2 )/(1+z^2 ))) ((2dz)/(1+z^2 )) =−2 ∫ ((arctan(z))/(1−z^2 )) dz.....be continued...
$$\mathrm{we}\:\mathrm{have}\:\int\:\:\frac{\mathrm{1}}{\mathrm{a}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\frac{\mathrm{1}−\mathrm{a}}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}\right)\mathrm{da} \\ $$$$=_{\mathrm{a}\:=\mathrm{cost}} \:\:\:\:\int\:\frac{\mathrm{1}}{\mathrm{cost}\:\mathrm{sint}}\:\mathrm{arctan}\left(\frac{\mathrm{1}−\mathrm{cost}}{\mathrm{sint}}\right)\left(−\mathrm{sint}\right)\mathrm{dt} \\ $$$$=−\int\:\:\frac{\mathrm{1}}{\mathrm{cost}}\:\mathrm{arctan}\left(\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{2sin}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}\right)\mathrm{dt} \\ $$$$=−\int\:\:\frac{\mathrm{1}}{\mathrm{cost}}\:\mathrm{arctan}\left(\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{t}}{\mathrm{cost}}\:\mathrm{dt}\right. \\ $$$$=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{z}} \:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{2arctan}\left(\mathrm{z}\right)}{\frac{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }}\:\frac{\mathrm{2dz}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} } \\ $$$$=−\mathrm{2}\:\int\:\:\:\frac{\mathrm{arctan}\left(\mathrm{z}\right)}{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:\mathrm{dz}…..\mathrm{be}\:\mathrm{continued}… \\ $$

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