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Question Number 85426 by john santu last updated on 22/Mar/20
∫ _(π/4)^0 ((sin x+cos x)/(9+16sin 2x)) dx
$$\int\:_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{9}+\mathrm{16sin}\:\mathrm{2}{x}}\:{dx} \\ $$
Commented by som(math1967) last updated on 22/Mar/20
∫((sinx+cosx)/(25−16(1−sin2x)))dx (1/4)∫((4(sinx+cosx))/(5^2 −(4sinx−4cosx)^2 ))dx (1/4)∫((d(4sinx−4cosx))/(5^2 −(4sinx−4cosx)^2 ))
$$\int\frac{{sinx}+{cosx}}{\mathrm{25}−\mathrm{16}\left(\mathrm{1}−{sin}\mathrm{2}{x}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{4}\left({sinx}+{cosx}\right)}{\mathrm{5}^{\mathrm{2}} −\left(\mathrm{4}{sinx}−\mathrm{4}{cosx}\right)^{\mathrm{2}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left(\mathrm{4}{sinx}−\mathrm{4}{cosx}\right)}{\mathrm{5}^{\mathrm{2}} −\left(\mathrm{4}{sinx}−\mathrm{4}{cosx}\right)^{\mathrm{2}} }\: \\ $$
Commented by niroj last updated on 22/Mar/20
∫_0 ^(π/4) (( sin x+ cos x)/(9+16sin2x))dx sin2x= 1+2sinxcos x−1 = 1+2sin x cos x−sin^2 x−cos^2 x = 1−(sin^2 x−2sinx cos x+cos^2 x) = 1−(sin x−cos x)^2 = ∫_0 ^(π/4) (((sin x+ cos x)dx)/(9+16{1−(sin x−cos x)^2 })) put, sin x−cos x=t (cos x+sinx)dx=dt if x=(π/4) ⇒ t=0 if x=0⇒ t=−1 ∫_(−1) ^( 0) (( dt)/(9+16(1−t^2 ))) = ∫_(−1) ^( 0) (( dt)/(9+16−16t^2 )) = ∫_(−1) ^0 (( 1)/(25−16t^2 ))dt = ∫_( −1) ^( 0) (( 1)/(16(((25)/(16))−t^2 )))dt =(1/(16)) ∫_(−1) ^( 0) (1/(((5/4))^2 −(t)^2 ))dt = (1/(16))[ (1/(2.(5/4)))log (((5/4)+t)/((5/4)−t))]_(−1) ^0 = (1/(16))[(2/5)log ((5+4t)/(5−4t))]_(−1) ^0 = (1/(16))[((2/5)log(5/5))−((2/5)log((5−4)/(5+4)))] = (1/(16))[(2/5)log 1−(2/5)log(1/9)] = (1/(16))[ (2/5).0−(2/5)log(1/3^2 )] = (1/(16))(0−(2/5)log3^(−2) =(1/(16))×(−(2/5))×(−2).log3 = ((4.1)/(16.5))log3 = ((log 3)/(20))//
$$\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\:\:\mathrm{sin}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{x}}{\mathrm{9}+\mathrm{16sin2x}}\mathrm{dx} \\ $$$$\:\:\:\:\mathrm{sin2x}=\:\mathrm{1}+\mathrm{2sinxcos}\:\mathrm{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}+\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{cos}^{\mathrm{2}} \mathrm{x} \\ $$$$\:\:\:\:\:=\:\mathrm{1}−\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{2sinx}\:\mathrm{cos}\:\mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right) \\ $$$$\:\:\:\:\:=\:\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \\ $$$$=\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\left(\mathrm{sin}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx}}{\mathrm{9}+\mathrm{16}\left\{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \right\}} \\ $$$$\:\mathrm{put},\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}=\mathrm{t} \\ $$$$\:\:\:\:\left(\mathrm{cos}\:\mathrm{x}+\mathrm{sinx}\right)\mathrm{dx}=\mathrm{dt} \\ $$$$\:\mathrm{if}\:\mathrm{x}=\frac{\pi}{\mathrm{4}}\:\Rightarrow\:\mathrm{t}=\mathrm{0} \\ $$$$\:\mathrm{if}\:\mathrm{x}=\mathrm{0}\Rightarrow\:\mathrm{t}=−\mathrm{1} \\ $$$$\:\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \:\:\frac{\:\:\mathrm{dt}}{\mathrm{9}+\mathrm{16}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)} \\ $$$$\:\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \:\:\frac{\:\:\:\mathrm{dt}}{\mathrm{9}+\mathrm{16}−\mathrm{16t}^{\mathrm{2}} } \\ $$$$=\:\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{\:\mathrm{1}}{\mathrm{25}−\mathrm{16t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\:\:\int_{\:−\mathrm{1}} ^{\:\mathrm{0}} \:\:\frac{\:\mathrm{1}}{\mathrm{16}\left(\frac{\mathrm{25}}{\mathrm{16}}−\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \:\frac{\mathrm{1}}{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\:\frac{\mathrm{1}}{\mathrm{2}.\frac{\mathrm{5}}{\mathrm{4}}}\mathrm{log}\:\frac{\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{t}}{\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{t}}\right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\:\frac{\mathrm{5}+\mathrm{4t}}{\mathrm{5}−\mathrm{4t}}\right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\left(\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\frac{\mathrm{5}}{\mathrm{5}}\right)−\left(\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\frac{\mathrm{5}−\mathrm{4}}{\mathrm{5}+\mathrm{4}}\right)\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\:\mathrm{1}−\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\frac{\mathrm{1}}{\mathrm{9}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\:\frac{\mathrm{2}}{\mathrm{5}}.\mathrm{0}−\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log}\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{0}−\frac{\mathrm{2}}{\mathrm{5}}\mathrm{log3}^{−\mathrm{2}} \right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}×\left(−\frac{\mathrm{2}}{\mathrm{5}}\right)×\left(−\mathrm{2}\right).\mathrm{log3} \\ $$$$=\:\frac{\mathrm{4}.\mathrm{1}}{\mathrm{16}.\mathrm{5}}\mathrm{log3}\: \\ $$$$=\:\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{20}}// \\ $$$$ \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$
Commented by mathmax by abdo last updated on 22/Mar/20
I =−∫_0 ^(π/4) ((sinx +cosx)/(9+32 sinx cosx))dx changement tan((x/2))=t give I =−∫_0 ^((√2)−1) ((((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 )))/(9+32((2t)/(1+t^2 ))×((1−t^2 )/(1+t^2 ))))((2dt)/(1+t^2 )) = −∫_0 ^((√2)−1) (1/((1+t^2 )^2 ))×((2t+1−t^2 )/((9+32((2t(1−t^2 ))/((1+t^2 )^2 )))))dt =∫_0 ^((√2)−1) ((t^2 −2t−1)/(9(1+t^2 )^2 +64t(1−t^2 )))dt =∫_0 ^((√2)−1) ((t^2 −2t−1)/(9(t^4 +2t^2 +1)+64t −64 t^3 ))dt =∫_0 ^((√2)−1) ((t^2 −2t−1)/(9t^4 −64t^3 +18t^2 +64t +9))dt =∫_0 ^((√2)−1) F(t)dt let decompose F(t) the roots of 9t^4 −64t^3 +18t^2 +64 t +9 =0 are t_1 ∼6,64 58 t_2 ∼1,3542 t_3 ∼ −0,7364 t_4 ∼−0,1505 ⇒F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +(c/(t−t_3 )) +(d/(t−t_4 )) ⇒ ∫ F(t)dt =aln∣t−t_1 ∣ +ln∣t−t_2 ∣ +cln∣t−t_3 ∣ +dln∣t−t_4 ∣ +c =aln∣tan((x/2))−t_1 ∣+ln∣tan((x/2))−t_2 ∣ +cln∣tan((x/2))−t_3 ∣ +dln∣tan((x/2))−t_4 ∣ +d rest calculus of coefficients...
$${I}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}\:+{cosx}}{\mathrm{9}+\mathrm{32}\:{sinx}\:{cosx}}{dx}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{9}+\mathrm{32}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{2}{t}+\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{9}+\mathrm{32}\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{\mathrm{9}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{64}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{\mathrm{9}\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{64}{t}\:−\mathrm{64}\:{t}^{\mathrm{3}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{\mathrm{9}{t}^{\mathrm{4}} −\mathrm{64}{t}^{\mathrm{3}} +\mathrm{18}{t}^{\mathrm{2}} \:+\mathrm{64}{t}\:+\mathrm{9}}{dt}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:{F}\left({t}\right){dt} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)\: \\ $$$${the}\:{roots}\:{of}\:\mathrm{9}{t}^{\mathrm{4}} −\mathrm{64}{t}^{\mathrm{3}} \:+\mathrm{18}{t}^{\mathrm{2}} \:+\mathrm{64}\:{t}\:+\mathrm{9}\:=\mathrm{0}\:{are} \\ $$$${t}_{\mathrm{1}} \sim\mathrm{6},\mathrm{64}\:\mathrm{58} \\ $$$${t}_{\mathrm{2}} \sim\mathrm{1},\mathrm{3542} \\ $$$${t}_{\mathrm{3}} \sim\:−\mathrm{0},\mathrm{7364} \\ $$$${t}_{\mathrm{4}} \sim−\mathrm{0},\mathrm{1505}\:\Rightarrow{F}\left({t}\right)=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{c}}{{t}−{t}_{\mathrm{3}} }\:+\frac{{d}}{{t}−{t}_{\mathrm{4}} }\:\Rightarrow \\ $$$$\int\:{F}\left({t}\right){dt}\:={aln}\mid{t}−{t}_{\mathrm{1}} \mid\:+{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+{cln}\mid{t}−{t}_{\mathrm{3}} \mid\:+{dln}\mid{t}−{t}_{\mathrm{4}} \mid\:+{c} \\ $$$$={aln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{1}} \mid+{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{2}} \mid\:+{cln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{3}} \mid \\ $$$$+{dln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{t}_{\mathrm{4}} \mid\:+{d}\:\:\:{rest}\:{calculus}\:{of}\:{coefficients}… \\ $$
Commented by mathmax by abdo last updated on 22/Mar/20
let try another way we[have I =−∫_0 ^(π/4) ((sinx +cosx)/(9+16sin(2x)))dx ⇒I =−∫_0 ^(π/4) (((√2)cos(x−(π/4)))/(9+16sin(2x)))dx =_(x−(π/4)=−t) −(√2)∫_(π/4) ^0 ((cos(t))/(9+16sin(2(−t+(π/4)))))dt =−(√2)∫_0 ^(π/4) ((cost)/(9+16 sin(−2t+(π/2))))dt =−(√2)∫_0 ^(π/4) ((cost dt)/(9+16 cos(2t))) =−(√2)∫_0 ^(π/4) ((cost dt)/(9+16(1−2sin^2 t))) =−(√2)∫_0 ^(π/4) ((cost dt)/(25−32 sin^2 t)) =_(sint =u) −(√2)∫_0 ^((√2)/2) (du/(25−32u^2 )) =(√2)∫_0 ^(1/( (√2))) (du/(32u^2 −25)) =((√2)/(32))∫_0 ^(1/( (√2))) (du/(u^2 −((25)/(32)))) =_(u=(5/( (√(32))))z) ((√2)/(32)) ∫_0 ^((√(32))/(5(√2))) (1/(((25)/(32))(z^2 −1)))(5/( (√(32))))dz =((√2)/(32))×((32)/(25))×(5/( (√(32)))) ∫_0 ^(4/5) (dz/(z^2 −1)) =((√2)/(10(√(32)))) ∫_0 ^(4/5) ((1/(z−1))−(1/(z+1)))dz =(1/(40)) [ln∣((z−1)/(z+1))∣]_0 ^(4/5) =(1/(40)){ln∣(((4/5)−1)/((4/5)+1))∣} =(1/(40))ln∣(1/9)∣ =−(1/(40))ln(3^2 ) =−((ln(3))/(20))
$${let}\:{try}\:{another}\:{way}\:{we}\left[{have}\right. \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}\:+{cosx}}{\mathrm{9}+\mathrm{16}{sin}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow{I}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{9}+\mathrm{16}{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=_{{x}−\frac{\pi}{\mathrm{4}}=−{t}} \:\:\:−\sqrt{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \:\:\frac{{cos}\left({t}\right)}{\mathrm{9}+\mathrm{16}{sin}\left(\mathrm{2}\left(−{t}+\frac{\pi}{\mathrm{4}}\right)\right)}{dt} \\ $$$$=−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{cost}}{\mathrm{9}+\mathrm{16}\:{sin}\left(−\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right)}{dt}\:=−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{cost}\:{dt}}{\mathrm{9}+\mathrm{16}\:{cos}\left(\mathrm{2}{t}\right)} \\ $$$$=−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{cost}\:{dt}}{\mathrm{9}+\mathrm{16}\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {t}\right)}\:=−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{cost}\:{dt}}{\mathrm{25}−\mathrm{32}\:{sin}^{\mathrm{2}} {t}} \\ $$$$=_{{sint}\:={u}} \:\:\:−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\frac{{du}}{\mathrm{25}−\mathrm{32}{u}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\:\frac{{du}}{\mathrm{32}{u}^{\mathrm{2}} −\mathrm{25}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{32}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{32}}}\:=_{{u}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{32}}}{z}} \:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{32}}}{\mathrm{5}\sqrt{\mathrm{2}}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{25}}{\mathrm{32}}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}\frac{\mathrm{5}}{\:\sqrt{\mathrm{32}}}{dz} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{32}}×\frac{\mathrm{32}}{\mathrm{25}}×\frac{\mathrm{5}}{\:\sqrt{\mathrm{32}}}\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\mathrm{5}}} \:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{10}\sqrt{\mathrm{32}}}\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\mathrm{5}}} \:\left(\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{1}}\right){dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}}\:\left[{ln}\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{4}}{\mathrm{5}}} \:=\frac{\mathrm{1}}{\mathrm{40}}\left\{{ln}\mid\frac{\frac{\mathrm{4}}{\mathrm{5}}−\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{5}}+\mathrm{1}}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}}{ln}\mid\frac{\mathrm{1}}{\mathrm{9}}\mid\:=−\frac{\mathrm{1}}{\mathrm{40}}{ln}\left(\mathrm{3}^{\mathrm{2}} \right)\:=−\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{20}} \\ $$
Answered by john santu last updated on 22/Mar/20

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