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Question Number 105862 by bemath last updated on 01/Aug/20
∫ (dx/(9+16cos^2 x)) ?
$$\int\:\frac{{dx}}{\mathrm{9}+\mathrm{16cos}\:^{\mathrm{2}} {x}}\:? \\ $$
Commented by bemath last updated on 01/Aug/20
thank you
$${thank}\:{you} \\ $$
Answered by Mr.D.N. last updated on 01/Aug/20
∫ (( dx)/(9+16cos^2 x))= ∫ (( (1/(cos^2 x)))/((9+16cos^2 x)/(cos^2 x)))dx ∫ (( sec^2 x dx)/(9sec^2 x+16))= ∫ ((sec^2 xdx)/(9(1+tan^2 x)+16)) = ∫ ((sec^2 xdx)/(9tan^2 x+25))= (1/9)∫ ((sec^2 xdx)/(tan^2 x+((25)/9))) Put, tan x= t sec^2 xdx=dt (1/9)∫ (( dt)/(t^2 +((5/3))^2 )) = (1/9)×(1/(5/3)) tan^(−1) (t/(5/3))+C = (3/(9×5)) tan^(−1) ((3t)/5)+C = (1/(15)) tan^(−1) ((3 tan x)/5)+C //.
$$\:\:\int\:\frac{\:\:\mathrm{dx}}{\mathrm{9}+\mathrm{16cos}^{\mathrm{2}} \mathrm{x}}=\:\int\:\frac{\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}}{\frac{\mathrm{9}+\mathrm{16cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\ $$$$\:\int\:\frac{\:\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\mathrm{9sec}^{\mathrm{2}} \mathrm{x}+\mathrm{16}}=\:\int\:\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\mathrm{9}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{16}} \\ $$$$\:=\:\int\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\mathrm{9tan}^{\mathrm{2}} \mathrm{x}+\mathrm{25}}=\:\frac{\mathrm{1}}{\mathrm{9}}\int\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{25}}{\mathrm{9}}} \\ $$$$\:\:\mathrm{Put},\:\mathrm{tan}\:\mathrm{x}=\:\mathrm{t} \\ $$$$\:\:\:\:\mathrm{sec}^{\mathrm{2}} \mathrm{xdx}=\mathrm{dt} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{9}}\int\:\frac{\:\:\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{9}}×\frac{\mathrm{1}}{\frac{\mathrm{5}}{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{t}}{\frac{\mathrm{5}}{\mathrm{3}}}+\mathrm{C} \\ $$$$\:\:=\:\:\frac{\mathrm{3}}{\mathrm{9}×\mathrm{5}}\:\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{3t}}{\mathrm{5}}+\mathrm{C} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{15}}\:\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{3}\:\mathrm{tan}\:\mathrm{x}}{\mathrm{5}}+\mathrm{C}\://. \\ $$
Answered by Dwaipayan Shikari last updated on 01/Aug/20
∫(dx/(9+16cos^2 x))=∫((sec^2 x)/(9sec^2 x+16))=∫((sec^2 x)/(9tan^2 x+25))dx ∫(dt/(9t^2 +25)) (tanx=t) (1/9)∫(dt/(t^2 +((5/3))^2 ))=(1/9).(3/5)tan^(−1) ((3t)/5)+C=(1/(15))tan^(−1) ((3tanx)/5)+C
$$\int\frac{{dx}}{\mathrm{9}+\mathrm{16}{cos}^{\mathrm{2}} {x}}=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{9}{sec}^{\mathrm{2}} {x}+\mathrm{16}}=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{9}{tan}^{\mathrm{2}} {x}+\mathrm{25}}{dx} \\ $$$$\int\frac{{dt}}{\mathrm{9}{t}^{\mathrm{2}} +\mathrm{25}}\:\:\:\:\:\:\left({tanx}={t}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{9}}.\frac{\mathrm{3}}{\mathrm{5}}{tan}^{−\mathrm{1}} \frac{\mathrm{3}{t}}{\mathrm{5}}+{C}=\frac{\mathrm{1}}{\mathrm{15}}{tan}^{−\mathrm{1}} \frac{\mathrm{3}{tanx}}{\mathrm{5}}+{C} \\ $$
Answered by mathmax by abdo last updated on 01/Aug/20
I =∫ (dx/(9+16cos^2 x)) ⇒ I =∫ (dx/(9+16×((1+cos(2x))/2))) =∫ (dx/(9+8 +8cos(2x))) =∫ (dx/(17+8cos(2x))) =_(2x=t) ∫ (dt/(2(17+8cost))) 2I=_(tan((t/2))=u) ∫ ((2du)/((1+u^2 )(17+8((1−u^2 )/(1+u^2 ))))) =∫ ((2du)/(17 +17u^2 +8−8u^2 )) =∫ ((2du)/(25+9u^2 )) =(2/(25)) ∫ (du/(1+(9/(25))u^2 )) =_(z =(3/5)u) (2/(25)) ∫ (1/(1+z^2 ))×(5/3)dz =(2/(15)) arctanz +C =(2/(15)) arctan(((3u)/5)) +C =(2/(25)) arctan((3/5) tan((t/2))) +C =(2/(15)) arctan((3/5)tan(x)) +C ⇒ I =(1/(15)) arctan((3/5)tanx) +C
$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{9}+\mathrm{16cos}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\:\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{9}+\mathrm{16}×\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}} \\ $$$$=\int\:\:\frac{\mathrm{dx}}{\mathrm{9}+\mathrm{8}\:+\mathrm{8cos}\left(\mathrm{2x}\right)}\:=\int\:\frac{\mathrm{dx}}{\mathrm{17}+\mathrm{8cos}\left(\mathrm{2x}\right)}\:=_{\mathrm{2x}=\mathrm{t}} \:\:\:\:\int\:\:\frac{\mathrm{dt}}{\mathrm{2}\left(\mathrm{17}+\mathrm{8cost}\right)} \\ $$$$\mathrm{2I}=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{u}} \:\:\:\int\:\:\frac{\mathrm{2du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\left(\mathrm{17}+\mathrm{8}\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\:\frac{\mathrm{2du}}{\mathrm{17}\:+\mathrm{17u}^{\mathrm{2}} \:+\mathrm{8}−\mathrm{8u}^{\mathrm{2}} }\:=\int\:\frac{\mathrm{2du}}{\mathrm{25}+\mathrm{9u}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{25}}\:\int\:\:\frac{\mathrm{du}}{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{25}}\mathrm{u}^{\mathrm{2}} } \\ $$$$=_{\mathrm{z}\:=\frac{\mathrm{3}}{\mathrm{5}}\mathrm{u}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{25}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }×\frac{\mathrm{5}}{\mathrm{3}}\mathrm{dz}\:=\frac{\mathrm{2}}{\mathrm{15}}\:\mathrm{arctanz}\:+\mathrm{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{15}}\:\mathrm{arctan}\left(\frac{\mathrm{3u}}{\mathrm{5}}\right)\:+\mathrm{C}\:=\frac{\mathrm{2}}{\mathrm{25}}\:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{15}}\:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tan}\left(\mathrm{x}\right)\right)\:+\mathrm{C}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{15}}\:\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tanx}\right)\:+\mathrm{C} \\ $$

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