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Question Number 58259 by behi83417@gmail.com last updated on 20/Apr/19

a .∫ (dx/(2sin^2 x+3tg^2 x))=? b .∫(( 1+(x)^(1/3) )/(1+(√x)+(x)^(1/3) +(x)^(1/6) ))dx=? c .∫ ((cosx)/(1+cos2x))dx=? d .∫ ((sin^2 x)/((√2)+(√3).cos^2 x))dx=?

$$\boldsymbol{\mathrm{a}}\:\:.\int\:\:\frac{\boldsymbol{\mathrm{dx}}}{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\mathrm{3}\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}=? \\ $$$$\boldsymbol{\mathrm{b}}\:\:\:.\int\frac{\:\:\mathrm{1}+\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}}{\mathrm{1}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}+\sqrt[{\mathrm{6}}]{\boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}}=? \\ $$$$\boldsymbol{\mathrm{c}}\:\:\:\:\:.\int\:\:\frac{\boldsymbol{\mathrm{cosx}}}{\mathrm{1}+\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=? \\ $$$$\boldsymbol{\mathrm{d}}\:\:\:\:\:.\int\:\:\:\frac{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}{\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}.\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=? \\ $$

Commented by maxmathsup by imad last updated on 21/Apr/19

yes sir i have forgotten (√3) but give opportonity to do method sometimes i feel tired because of the work thanks..

$${yes}\:{sir}\:{i}\:{have}\:{forgotten}\:\sqrt{\mathrm{3}}\:\:{but}\:{give}\:{opportonity}\:{to}\:{do}\:{method}\:{sometimes} \\ $$$${i}\:{feel}\:{tired}\:\:{because}\:{of}\:{the}\:{work}\:\:{thanks}.. \\ $$

Commented by behi83417@gmail.com last updated on 21/Apr/19

dear abdo! thank you for so hard and nice work. excuse me sir.something went wrong in line#3 from above. ⇒2(√2)+(√3)+(√3).cost ,is right.

$${dear}\:{abdo}!\:{thank}\:{you}\:{for}\:{so}\:{hard}\:{and}\:{nice}\:{work}. \\ $$$${excuse}\:{me}\:{sir}.{something}\:{went}\:{wrong}\:{in} \\ $$$${line}#\mathrm{3}\:{from}\:{above}. \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}.{cost}\:,{is}\:{right}. \\ $$

Commented by maxmathsup by imad last updated on 20/Apr/19

c) let I =∫ ((cosx)/(1+cos(2x)))dx ⇒ I =∫ ((cosx)/(1+2cos^2 x −1)) dx =∫ ((cosx)/(2cos^2 x)) dx =(1/2) ∫ (dx/(cosx)) =_(tan((x/2))=t) (1/2) ∫ ((2dt)/((1+t^2 )((1−t^2 )/(1+t^2 )))) = ∫ (dt/(1−t^2 )) =∫ ((1/(1−t)) +(1/(1+t)))dt =ln∣((1+t)/(1−t))∣ +c =ln∣((1+tan((x/2)))/(1−tan((x/2))))∣ +c

$$\left.{c}\right)\:{let}\:{I}\:=\int\:\:\frac{{cosx}}{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{I}\:=\int\:\:\:\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cos}^{\mathrm{2}} {x}\:−\mathrm{1}}\:{dx} \\ $$$$=\int\:\:\:\:\:\frac{{cosx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}}\:{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=\:\int\:\:\:\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}\:={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c}\:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{c} \\ $$

Commented by maxmathsup by imad last updated on 21/Apr/19

let A =∫ ((sin^2 x)/((√2) +(√3)cos^2 x))dx =∫ (((1−cos(2x))/2)/((√2) +(√3)((1+cos(2x))/2))) dx =∫ ((1−cos(2x))/(2(√2) +(√3) +(√3)cos(2x)))dx =_(2x =t) ∫ ((1−cos(t))/(2(√2) +(√3)+(√3)cos(t))) (dt/2) =(1/2) ∫ ((1−cos(t))/(2(√2) +(√3)cost)) dt =_(tan((t/2))=u) (1/2) ∫ ((1−((1−u^2 )/(1+u^2 )))/(2(√2) +(√3)((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫ ((2u^2 )/((1+u^2 )(2(√2)(1+u^2 )+(√3)−(√3)u^2 ))) du =∫ ((2u^2 )/((1+u^2 )(2(√2) +(√3) +(2(√2)−(√3))u^2 ))) du let decompose F(u) =((2u^2 )/((u^2 +1){au^2 +b})) (with a=2(√2)−(√3) and b=2(√2) +(√3)) F(u) = ((αu +β)/(u^2 +1)) +((cu +d)/(au^2 +b)) we have F(−u)=F(u) ⇒ ((−αu +β)/(u^2 +1)) +((−cu +d)/(au^2 +b)) =F(u) ⇒α=c =0 ⇒F(u) =(β/(u^2 +1)) +(d/(au^2 +b)) F(0) =0=β +(d/b) ⇒bβ +d =0 lim_(u→0) u^2 F(u) =(2/a) = β +(d/a) ⇒2 =βa +d ⇒d=2−βa ⇒ bβ +2−βa =0 ⇒(b−a)β =−2 ⇒β =(2/(a−b)) d =−bβ =((−2b)/(a−b)) ⇒F(u) =(2/((a−b)(u^2 +1))) −((2b)/((a−b)(au^2 +b))) ⇒ ∫ F(u)du =(2/(a−b)) ∫ (du/(u^2 +1)) −((2b)/((a−b))) ∫ (du/(au^2 +b)) ∫ (du/(u^2 +1)) =arctan(u) +c_0 ∫ (du/(au^2 +b)) =(1/a) ∫ (du/(u^2 +(b/a))) =_(u=(√(b/a))u) (1/a) ∫ (du/((b/a)(1 +u^2 ))) (√(b/a))du =(1/(√(ab))) arctan(u) +c_1 ⇒ A =(2/(a−b)) arctan(tanx) +(1/(√(ab))) actan(tanx) +c =((2x)/(a−b)) +(x/(√(ab))) =((2x)/(−2(√3))) +(x/(.(√5))) =((1/(√5)) −(1/(√3)))x

$${let}\:{A}\:=\int\:\frac{{sin}^{\mathrm{2}} {x}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}{cos}^{\mathrm{2}} {x}}{dx}\:=\int\:\:\:\frac{\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}\:{dx} \\ $$$$=\int\:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{3}}{cos}\left(\mathrm{2}{x}\right)}{dx}\:=_{\mathrm{2}{x}\:={t}} \:\:\:\int\:\:\frac{\mathrm{1}−{cos}\left({t}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}{cos}\left({t}\right)}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{1}−{cos}\left({t}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}{cost}}\:{dt}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)+\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{u}^{\mathrm{2}} \right)}\:{du} \\ $$$$=\int\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){u}^{\mathrm{2}} \right)}\:{du}\:\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{{au}^{\mathrm{2}} \:+{b}\right\}}\:\:\:\:\:\left({with}\:{a}=\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\:\:{and}\:\:{b}=\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right) \\ $$$${F}\left({u}\right)\:=\:\frac{\alpha{u}\:+\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{cu}\:+{d}}{{au}^{\mathrm{2}} \:+{b}}\:\:\:{we}\:{have}\:{F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow \\ $$$$\frac{−\alpha{u}\:+\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{au}^{\mathrm{2}} \:+{b}}\:={F}\left({u}\right)\:\Rightarrow\alpha={c}\:=\mathrm{0}\:\Rightarrow{F}\left({u}\right)\:=\frac{\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{d}}{{au}^{\mathrm{2}} \:+{b}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}=\beta\:\:+\frac{{d}}{{b}}\:\Rightarrow{b}\beta\:+{d}\:=\mathrm{0} \\ $$$${lim}_{{u}\rightarrow\mathrm{0}} \:\:\:{u}^{\mathrm{2}} {F}\left({u}\right)\:=\frac{\mathrm{2}}{{a}}\:=\:\beta\:+\frac{{d}}{{a}}\:\Rightarrow\mathrm{2}\:=\beta{a}\:+{d}\:\Rightarrow{d}=\mathrm{2}−\beta{a}\:\Rightarrow \\ $$$${b}\beta\:+\mathrm{2}−\beta{a}\:=\mathrm{0}\:\Rightarrow\left({b}−{a}\right)\beta\:=−\mathrm{2}\:\Rightarrow\beta\:=\frac{\mathrm{2}}{{a}−{b}} \\ $$$${d}\:=−{b}\beta\:=\frac{−\mathrm{2}{b}}{{a}−{b}}\:\Rightarrow{F}\left({u}\right)\:=\frac{\mathrm{2}}{\left({a}−{b}\right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:−\frac{\mathrm{2}{b}}{\left({a}−{b}\right)\left({au}^{\mathrm{2}} \:+{b}\right)}\:\Rightarrow \\ $$$$\int\:{F}\left({u}\right){du}\:=\frac{\mathrm{2}}{{a}−{b}}\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{2}{b}}{\left({a}−{b}\right)}\:\int\:\:\frac{{du}}{{au}^{\mathrm{2}} \:+{b}} \\ $$$$\int\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:={arctan}\left({u}\right)\:+{c}_{\mathrm{0}} \\ $$$$\int\:\:\:\frac{{du}}{{au}^{\mathrm{2}} \:+{b}}\:=\frac{\mathrm{1}}{{a}}\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{{b}}{{a}}}\:=_{{u}=\sqrt{\frac{{b}}{{a}}}{u}} \:\:\:\frac{\mathrm{1}}{{a}}\:\int\:\:\frac{{du}}{\frac{{b}}{{a}}\left(\mathrm{1}\:+{u}^{\mathrm{2}} \right)}\:\sqrt{\frac{{b}}{{a}}}{du} \\ $$$$=\frac{\mathrm{1}}{\sqrt{{ab}}}\:{arctan}\left({u}\right)\:+{c}_{\mathrm{1}} \:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}}{{a}−{b}}\:{arctan}\left({tanx}\right)\:+\frac{\mathrm{1}}{\sqrt{{ab}}}\:{actan}\left({tanx}\right)\:+{c} \\ $$$$=\frac{\mathrm{2}{x}}{{a}−{b}}\:+\frac{{x}}{\sqrt{{ab}}}\:=\frac{\mathrm{2}{x}}{−\mathrm{2}\sqrt{\mathrm{3}}}\:+\frac{{x}}{.\sqrt{\mathrm{5}}}\:=\left(\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right){x} \\ $$$$ \\ $$

Answered by tanmay last updated on 20/Apr/19

c)∫((cosx)/(2cos^2 x))dx (1/2)∫secxdx (1/2)ln(secx+tanx)+c

$$\left.{c}\right)\int\frac{{cosx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{secxdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({secx}+{tanx}\right)+{c} \\ $$

Answered by tanmay last updated on 20/Apr/19

b)x=t^6 →dx=6t^5 dt ∫((1+t^2 )/(1+t^3 +t^2 +t))×6t^5 dt 6∫((t^7 +t^5 )/(t^3 +t^2 +t+1))dt =6∫((t^7 +t^6 +t^5 +t^4 −t^6 −t^4 )/(t^3 +t^2 +t+1))dt =6∫t^4 −6∫((t^6 +t^4 )/(t^2 (t+1)+1(t^2 +1)))dt 6∫t^4 −6∫((t^4 (t^2 +1))/((t^2 +1)(t+1)))dt =6∫t^4 −6∫((t^4 −1+1)/(t+1))dt 6∫t^4 −6∫(((t^2 +1)(t−1)(t+1)+1)/(t+1))dt 6∫t^4 −6∫(t^2 +1)(t−1)dt−6∫(dt/(t+1)) 6∫t^4 −6∫t^3 −t^2 +t−1dt−6∫(dt/(t+1)) =((6t^5 )/5)−((6t^4 )/4)+((6t^3 )/3)−((6t^2 )/2)+6t−6ln(t+1)+c =(6/5)(x)^(5/6) −(3/2)(x)^(4/6) +2(x)^(3/6) −3(x)^(2/6) +6(x)^(1/6) −6ln(x^(1/6) +1)+c

$$\left.{b}\right){x}={t}^{\mathrm{6}} \:\rightarrow{dx}=\mathrm{6}{t}^{\mathrm{5}} {dt} \\ $$$$\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}}×\mathrm{6}{t}^{\mathrm{5}} {dt} \\ $$$$\mathrm{6}\int\frac{{t}^{\mathrm{7}} +{t}^{\mathrm{5}} }{{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt} \\ $$$$=\mathrm{6}\int\frac{{t}^{\mathrm{7}} +{t}^{\mathrm{6}} +{t}^{\mathrm{5}} +{t}^{\mathrm{4}} −{t}^{\mathrm{6}} −{t}^{\mathrm{4}} }{{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt} \\ $$$$=\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int\frac{{t}^{\mathrm{6}} +{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)+\mathrm{1}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$$\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int\frac{{t}^{\mathrm{4}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt} \\ $$$$=\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int\frac{{t}^{\mathrm{4}} −\mathrm{1}+\mathrm{1}}{{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)+\mathrm{1}}{{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}−\mathrm{1}\right){dt}−\mathrm{6}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$\mathrm{6}\int{t}^{\mathrm{4}} −\mathrm{6}\int{t}^{\mathrm{3}} −{t}^{\mathrm{2}} +{t}−\mathrm{1}{dt}−\mathrm{6}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$=\frac{\mathrm{6}{t}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{6}{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{6}{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{6}{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6}{t}−\mathrm{6}{ln}\left({t}+\mathrm{1}\right)+{c} \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left({x}\right)^{\frac{\mathrm{5}}{\mathrm{6}}} −\frac{\mathrm{3}}{\mathrm{2}}\left({x}\right)^{\frac{\mathrm{4}}{\mathrm{6}}} +\mathrm{2}\left({x}\right)^{\frac{\mathrm{3}}{\mathrm{6}}} −\mathrm{3}\left({x}\right)^{\frac{\mathrm{2}}{\mathrm{6}}} +\mathrm{6}\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} −\mathrm{6}{ln}\left({x}^{\frac{\mathrm{1}}{\mathrm{6}}} +\mathrm{1}\right)+{c} \\ $$

Answered by tanmay last updated on 20/Apr/19

a)∫(dx/(1−cos2x+3tan^2 x)) t=tanx→dt=sec^2 xdx dx=(dt/(1+t^2 )) ∫(dt/((1+t^2 )(1−((1−t^2 )/(1+t^2 ))+3t^2 ))) ∫(dt/(1+t^2 −1+t^2 +3t^2 +3t^4 )) ∫(dt/(3t^4 +5t^2 )) ∫(dt/(t^2 (3t^2 +5))) (1/5)∫((3t^2 +5−3t^2 )/(t^2 (3t^2 +5)))dt (1/5)∫(dt/t^2 )−(3/5)∫(dt/(3(t^2 +(5/(3 ))))) (1/5)×(t^(−2+1) /(−2+1))−(1/5)×(1/(√(5/3)))×tan^(−1) ((t/(√(5/3))))+c ((−1)/(5t))−((√3)/(5(√5)))tan^(−1) ((t/(√(5/3))))+c ((−1)/(5tanx))−((√3)/(5(√5)))tan^(−1) (((tanx)/(√(5/3))))+c

$$\left.{a}\right)\int\frac{{dx}}{\mathrm{1}−{cos}\mathrm{2}{x}+\mathrm{3}{tan}^{\mathrm{2}} {x}} \\ $$$${t}={tanx}\rightarrow{dt}={sec}^{\mathrm{2}} {xdx} \\ $$$${dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}{t}^{\mathrm{2}} \right)} \\ $$$$\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}^{\mathrm{4}} } \\ $$$$\int\frac{{dt}}{\mathrm{3}{t}^{\mathrm{4}} +\mathrm{5}{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}} \left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{5}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{5}−\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{5}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dt}}{{t}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{5}}\int\frac{{dt}}{\mathrm{3}\left({t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{3}\:}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}×\frac{{t}^{−\mathrm{2}+\mathrm{1}} }{−\mathrm{2}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}}×{tan}^{−\mathrm{1}} \left(\frac{{t}}{\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}}\right)+{c} \\ $$$$\frac{−\mathrm{1}}{\mathrm{5}{t}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{5}\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}}\right)+{c} \\ $$$$\frac{−\mathrm{1}}{\mathrm{5}{tanx}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{5}\sqrt{\mathrm{5}}}{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}}\right)+{c} \\ $$

Answered by tanmay last updated on 20/Apr/19

4)∫((sin^2 x)/(a+bcos^2 x))dx ∫(((1−cos2x)/2)/(a+b(((1+cos2x)/2))))dx ∫((1−cos2x)/(2a+b+bcos2x))dx (1/b)∫((1−cos2x)/((((2a+b)/b))+cos2x))dx (1/b)∫((1+(((2a+b)/b))−(((2a+b)/b))−cos2x)/((((2a+b)/b))+cos2x))dx let k=((2a+b)/b) (1/(b ))∫((1+k)/(k+cos2x))dx−(1/b)∫dx ((1+k)/b)∫(dx/(k+((1−tan^2 x)/(1+tan^2 x))))−(1/b)∫dx ((1+k)/b)∫((sec^2 xdx)/((k+1)+(k−1)tan^2 x))−(1/b)∫dx ((1+k)/b)×(1/(k−1))∫((d(tanx))/(((√((k+1)/(k−1))) )^2 +tan^2 x))−(1/b)∫dx (1/b)×((k+1)/(k−1))×(1/(√((k+1)/(k−1))))tan^(−1) (((tanx)/(√((k+1)/(k−1)))))−(1/b)x+c now a=(√2) b=(√3) k=((2a+b)/b)→k=((2(√2) +(√3))/(√3)) (1/(√3))×(√((((2(√2) +(√(3 )))/((√3) ))+1)/(((2(√2) +(√3))/(√3))−1))) ×tan^(−1) (((tanx)/(√(((√3) +(√2) )/(√2)))))−(1/(√3))x+c (1/(√3))×(√(((√3) +(√2) )/(√2))) ×tan^(−1) (((tanx)/(√(((√3) +(√2) )/(√2)))))−(1/(√3))x+c

$$\left.\mathrm{4}\right)\int\frac{{sin}^{\mathrm{2}} {x}}{{a}+{bcos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}}{{a}+{b}\left(\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}\right)}{dx} \\ $$$$\int\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}{a}+{b}+{bcos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{{b}}\int\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\left(\frac{\mathrm{2}{a}+{b}}{{b}}\right)+{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{{b}}\int\frac{\mathrm{1}+\left(\frac{\mathrm{2}{a}+{b}}{{b}}\right)−\left(\frac{\mathrm{2}{a}+{b}}{{b}}\right)−{cos}\mathrm{2}{x}}{\left(\frac{\mathrm{2}{a}+{b}}{{b}}\right)+{cos}\mathrm{2}{x}}{dx} \\ $$$${let}\:{k}=\frac{\mathrm{2}{a}+{b}}{{b}} \\ $$$$\frac{\mathrm{1}}{{b}\:}\int\frac{\mathrm{1}+{k}}{{k}+{cos}\mathrm{2}{x}}{dx}−\frac{\mathrm{1}}{{b}}\int{dx} \\ $$$$\frac{\mathrm{1}+{k}}{{b}}\int\frac{{dx}}{{k}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}−\frac{\mathrm{1}}{{b}}\int{dx} \\ $$$$\frac{\mathrm{1}+{k}}{{b}}\int\frac{{sec}^{\mathrm{2}} {xdx}}{\left({k}+\mathrm{1}\right)+\left({k}−\mathrm{1}\right){tan}^{\mathrm{2}} {x}}−\frac{\mathrm{1}}{{b}}\int{dx} \\ $$$$\frac{\mathrm{1}+{k}}{{b}}×\frac{\mathrm{1}}{{k}−\mathrm{1}}\int\frac{{d}\left({tanx}\right)}{\left(\sqrt{\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}}\:\right)^{\mathrm{2}} +{tan}^{\mathrm{2}} {x}}−\frac{\mathrm{1}}{{b}}\int{dx} \\ $$$$\frac{\mathrm{1}}{{b}}×\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}×\frac{\mathrm{1}}{\sqrt{\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}}}{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}}}\right)−\frac{\mathrm{1}}{{b}}{x}+{c} \\ $$$${now}\:{a}=\sqrt{\mathrm{2}}\:\:\:\:{b}=\sqrt{\mathrm{3}}\:\:\:{k}=\frac{\mathrm{2}{a}+{b}}{{b}}\rightarrow{k}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\sqrt{\frac{\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}\:}}{\sqrt{\mathrm{3}}\:}+\mathrm{1}}{\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}−\mathrm{1}}}\:\:×{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}{\sqrt{\mathrm{2}}}}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}{x}+{c} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\sqrt{\frac{\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}{\sqrt{\mathrm{2}}}}\:×{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\sqrt{\frac{\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}{\sqrt{\mathrm{2}}}}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}{x}+{c} \\ $$$$ \\ $$

Commented by behi83417@gmail.com last updated on 20/Apr/19

sir tanmay!thank you very much for so hard work.nice and smart.

$${sir}\:{tanmay}!{thank}\:{you}\:{very}\:{much}\:{for} \\ $$$${so}\:{hard}\:{work}.{nice}\:{and}\:{smart}. \\ $$

Commented by peter frank last updated on 20/Apr/19

nice work.thanks

$${nice}\:{work}.{thanks} \\ $$

Commented by tanmay last updated on 21/Apr/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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