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Question Number 84766 by jagoll last updated on 15/Mar/20
∫ (dx/((16+9sin x)^2 ))
$$\int\:\frac{\mathrm{dx}}{\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by jagoll last updated on 16/Mar/20
∫ sec x [ ((cos x)/((16+9sin x)^2 ))] dx = u = sec x ⇒ du = sec x tan x dx v = (1/9)∫ ((d(16+9sin x))/((16+9sin x)^2 )) = −(1/(16+9sin x)) ⇒ −((sec x)/(16+9sin x)) + ∫ ((sec x tan x)/(16+9sin x)) dx I_2 = ∫ ((sin x dx)/(cos^2 x (16+9sin x)))
$$\int\:\mathrm{sec}\:\:\mathrm{x}\:\left[\:\frac{\mathrm{cos}\:\:\mathrm{x}}{\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)^{\mathrm{2}} }\right]\:\mathrm{dx}\:= \\ $$$$ \\ $$$$\mathrm{u}\:=\:\mathrm{sec}\:\mathrm{x}\:\Rightarrow\:\mathrm{du}\:=\:\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{v}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\int\:\:\frac{\mathrm{d}\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)}{\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{16}+\mathrm{9sin}\:\mathrm{x}} \\ $$$$\Rightarrow\:−\frac{\mathrm{sec}\:\mathrm{x}}{\mathrm{16}+\mathrm{9sin}\:\mathrm{x}}\:+\:\int\:\:\frac{\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}{\mathrm{16}+\mathrm{9sin}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\mathrm{I}_{\mathrm{2}} \:=\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)} \\ $$
Commented by abdomathmax last updated on 16/Mar/20
let f(a) =∫ (dx/(a+9sinx)) we have f^′ (a)=−∫ (dx/((a+9sinx)^2 )) ⇒∫ (dx/((a +9sinx)^2 )) =−f^′ (a) f(a)=_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )(a+9×((2t)/(1+t^2 ))))) =∫ ((2dt)/(a+at^2 +18t)) =∫ ((2dt)/(at^2 +18t +a)) Δ^′ =9^2 −a^2 we take a>9 ⇒t_1 =((−9+(√(81−a^2 )))/a) t_2 =((−9−(√(81−a^2 )))/a) ⇒f(a)=∫ ((2dt)/(a(t−t_1 )(t−t_2 ))) =(2/(a×((2(√(81−a^2 )))/a)))∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/( (√(81−a^2 ))))ln∣((t−t_1 )/(t−t_2 ))∣ +C =(1/( (√(81−a^2 ))))ln∣((t−((−9+(√(81−a^2 )))/a))/(t−((−9−(√(81−a^2 )))/a)))∣ +C =(1/( (√(81−a^2 ))))ln∣((at+9−(√(81−a^2 )))/(at+9+(√(81−a^2 ))))∣ +C rest calculus of f^′ (a)...be continued...
$${let}\:{f}\left({a}\right)\:=\int\:\:\frac{{dx}}{{a}+\mathrm{9}{sinx}}\:\:{we}\:{have}\:{f}^{'} \left({a}\right)=−\int\:\:\frac{{dx}}{\left({a}+\mathrm{9}{sinx}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int\:\:\frac{{dx}}{\left({a}\:+\mathrm{9}{sinx}\right)^{\mathrm{2}} }\:=−{f}^{'} \left({a}\right) \\ $$$${f}\left({a}\right)=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+\mathrm{9}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\frac{\mathrm{2}{dt}}{{a}+{at}^{\mathrm{2}} \:+\mathrm{18}{t}}\:=\int\:\frac{\mathrm{2}{dt}}{{at}^{\mathrm{2}} \:+\mathrm{18}{t}\:+{a}} \\ $$$$\Delta^{'} =\mathrm{9}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:{we}\:{take}\:{a}>\mathrm{9}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{9}+\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{a}} \\ $$$${t}_{\mathrm{2}} =\frac{−\mathrm{9}−\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{a}}\:\Rightarrow{f}\left({a}\right)=\int\:\:\frac{\mathrm{2}{dt}}{{a}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{{a}×\frac{\mathrm{2}\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{a}}}\int\:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{ln}\mid\frac{{t}−\frac{−\mathrm{9}+\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{a}}}{{t}−\frac{−\mathrm{9}−\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{a}}}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{ln}\mid\frac{{at}+\mathrm{9}−\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}{{at}+\mathrm{9}+\sqrt{\mathrm{81}−{a}^{\mathrm{2}} }}\mid\:+{C} \\ $$$${rest}\:{calculus}\:{of}\:{f}^{'} \left({a}\right)…{be}\:{continued}… \\ $$
Answered by TANMAY PANACEA last updated on 16/Mar/20
t=a+bsinx→sinx=((t−a)/b) p=((cosx)/(a+bsinx)) (dp/dx)=(((a+bsinx)×−sinx−cosx(bcosx))/((a+bsinx)^2 )) (dp/dx)=((−asinx−bsin^2 x−bcos^2 x)/((a+bsinx)^2 ))=((−(asinx+b))/((a+bsinx)^2 )) now putting sinx=((t−a)/b) −(dp/dx)=(((a×((t−a)/b)+b))/t^2 )=((at−a^2 +b^2 )/(bt^2 ))=(a/b)×(1/t)+((b^2 −a^2 )/t^2 ) −(dp/dx)=(a/b)×(1/t)+((b^2 −a^2 )/t^2 ) ((−1)/((b^2 −a^2 )))(dp/dx)=(a/(b(b^2 −a^2 )))×(1/((a+bsinx)))+(1/((a+bsinx)^2 )) (1/(a^2 −b^2 ))×d(((cosx)/(a+bsinx)))=(a/(b(b^2 −a^2 )))×(dx/((a+bsinx)))+(dx/((a+bsinx)^2 )) now ∫(dx/((a+bsinx)^2 )) =(1/(a^2 −b^2 ))∫d(((cosx)/(a+bsinx)))−(a/(b(b^2 −a^2 )))∫(dx/(a+bsinx)) a=16 b=9 ∫(dx/((16+9sinx)^2 ))=(1/((16^2 −9^2 )))∫d(((cosx)/(16+9sinx)))+((16)/(9(16^2 −9^2 )))∫(dx/(16+9sinx)) now ∫(dx/(16+9sinx)) k=tan(x/2)→dk=sec^2 (x/2)×(1/2)dx→((2dk)/(1+k^2 ))=dx ∫((2dk)/((1+k^2 )))×(1/((16+9×((2k)/(1+k^2 ))))) 2∫(dk/(16+16k^2 +18k))=(1/8)∫(dk/(k^2 +((9k)/8)+1))=(1/8)∫(dk/(k^2 +2.k.(9/(16))+((9/(16)))^2 +1−((9/(16)))^2 )) =(1/8)∫(dk/((k+(9/(16)))^2 +((175)/(16^2 ))))=(1/8)×∫(dk/((k+(9/(16)))^2 +(((5(√7))/(16)))^2 )) =(1/8)×((16)/(5(√7)))tan^(−1) (((k+(9/(16)))/((5(√7))/(16))))=(2/(5(√7)))tan^(−1) (((16k+9)/(5(√7)))) so answer is (1/((16^2 −9^2 )))×(((cosx)/(16+9sinx)))+((16)/(9(16^2 −9^2 )))×(2/(5(√7)))tan^(−1) (((16tan(x/2)+9)/(5(√7))))+c
$${t}={a}+{bsinx}\rightarrow{sinx}=\frac{{t}−{a}}{{b}} \\ $$$${p}=\frac{{cosx}}{{a}+{bsinx}} \\ $$$$\frac{{dp}}{{dx}}=\frac{\left({a}+{bsinx}\right)×−{sinx}−{cosx}\left({bcosx}\right)}{\left({a}+{bsinx}\right)^{\mathrm{2}} } \\ $$$$\frac{{dp}}{{dx}}=\frac{−{asinx}−{bsin}^{\mathrm{2}} {x}−{bcos}^{\mathrm{2}} {x}}{\left({a}+{bsinx}\right)^{\mathrm{2}} }=\frac{−\left({asinx}+{b}\right)}{\left({a}+{bsinx}\right)^{\mathrm{2}} } \\ $$$${now}\:{putting}\:{sinx}=\frac{{t}−{a}}{{b}} \\ $$$$−\frac{{dp}}{{dx}}=\frac{\left({a}×\frac{{t}−{a}}{{b}}+{b}\right)}{{t}^{\mathrm{2}} }=\frac{{at}−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{bt}^{\mathrm{2}} }=\frac{{a}}{{b}}×\frac{\mathrm{1}}{{t}}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} } \\ $$$$−\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}×\frac{\mathrm{1}}{\boldsymbol{{t}}}+\frac{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{t}}^{\mathrm{2}} } \\ $$$$\frac{−\mathrm{1}}{\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}×\frac{\mathrm{1}}{\left(\boldsymbol{{a}}+\boldsymbol{{bsinx}}\right)}+\frac{\mathrm{1}}{\left(\boldsymbol{{a}}+\boldsymbol{{bsinx}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }×\boldsymbol{{d}}\left(\frac{\boldsymbol{{cosx}}}{\boldsymbol{{a}}+\boldsymbol{{b}}{sinx}}\right)=\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}×\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{a}}+\boldsymbol{{bsinx}}\right)}+\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{a}}+\boldsymbol{{bsinx}}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{now}}\:\int\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\int{d}\left(\frac{{cosx}}{{a}+{bsinx}}\right)−\frac{{a}}{{b}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}\int\frac{{dx}}{{a}+{bsinx}} \\ $$$${a}=\mathrm{16}\:\:\:{b}=\mathrm{9} \\ $$$$\int\frac{{dx}}{\left(\mathrm{16}+\mathrm{9}{sinx}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{16}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} \right)}\int{d}\left(\frac{{cosx}}{\mathrm{16}+\mathrm{9}{sinx}}\right)+\frac{\mathrm{16}}{\mathrm{9}\left(\mathrm{16}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} \right)}\int\frac{{dx}}{\mathrm{16}+\mathrm{9}{sinx}} \\ $$$${now} \\ $$$$\int\frac{{dx}}{\mathrm{16}+\mathrm{9}{sinx}}\:\:{k}={tan}\frac{{x}}{\mathrm{2}}\rightarrow{dk}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{dx}\rightarrow\frac{\mathrm{2}{dk}}{\mathrm{1}+{k}^{\mathrm{2}} }={dx} \\ $$$$\int\frac{\mathrm{2}{dk}}{\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}×\frac{\mathrm{1}}{\left(\mathrm{16}+\mathrm{9}×\frac{\mathrm{2}{k}}{\mathrm{1}+{k}^{\mathrm{2}} }\right)} \\ $$$$\mathrm{2}\int\frac{{dk}}{\mathrm{16}+\mathrm{16}{k}^{\mathrm{2}} +\mathrm{18}{k}}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dk}}{{k}^{\mathrm{2}} +\frac{\mathrm{9}{k}}{\mathrm{8}}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dk}}{{k}^{\mathrm{2}} +\mathrm{2}.{k}.\frac{\mathrm{9}}{\mathrm{16}}+\left(\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} +\mathrm{1}−\left(\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dk}}{\left({k}+\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} +\frac{\mathrm{175}}{\mathrm{16}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{8}}×\int\frac{{dk}}{\left({k}+\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{16}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{16}}{\mathrm{5}\sqrt{\mathrm{7}}}{tan}^{−\mathrm{1}} \left(\frac{{k}+\frac{\mathrm{9}}{\mathrm{16}}}{\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{16}}}\right)=\frac{\mathrm{2}}{\mathrm{5}\sqrt{\mathrm{7}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{16}{k}+\mathrm{9}}{\mathrm{5}\sqrt{\mathrm{7}}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{16}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} \right)}×\left(\frac{{cosx}}{\mathrm{16}+\mathrm{9}{sinx}}\right)+\frac{\mathrm{16}}{\mathrm{9}\left(\mathrm{16}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} \right)}×\frac{\mathrm{2}}{\mathrm{5}\sqrt{\mathrm{7}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{16}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{9}}{\mathrm{5}\sqrt{\mathrm{7}}}\right)+{c} \\ $$$$ \\ $$

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