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Question Number 118338 by benjo_mathlover last updated on 17/Oct/20

solve ∫ (dx/(3−5sin x))

$$\:\:\:\:{solve}\:\int\:\frac{{dx}}{\mathrm{3}−\mathrm{5sin}\:{x}}\: \\ $$

Answered by Dwaipayan Shikari last updated on 17/Oct/20

∫(dx/(3−5sinx)) =2∫(dt/(3−((10t)/((1+t^2 ))))).(1/(1+t^2 )) t=tan(x/2) =2∫(dt/(3+3t^2 −10t))=(2/3)∫(dt/((t−(5/3))^2 +1−((25)/9))) =(2/3)∫(dt/((t−(5/3))^2 −((4/3))^2 )) =(1/4)log(((t−(5/3)−(4/3))/(t−(5/3)+(4/3))))=(1/4)log(((3t−9)/(3t−1)))=(1/4)log(((3tan(x/2)−9)/(3tan(x/2)−1)))+4

$$\int\frac{{dx}}{\mathrm{3}−\mathrm{5}{sinx}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{3}−\frac{\mathrm{10}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} −\mathrm{10}{t}}=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dt}}{\left({t}−\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{25}}{\mathrm{9}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dt}}{\left({t}−\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{log}\left(\frac{{t}−\frac{\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}}{{t}−\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\mathrm{4}}{log}\left(\frac{\mathrm{3}{t}−\mathrm{9}}{\mathrm{3}{t}−\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}}{log}\left(\frac{\mathrm{3}{tan}\frac{{x}}{\mathrm{2}}−\mathrm{9}}{\mathrm{3}{tan}\frac{{x}}{\mathrm{2}}−\mathrm{1}}\right)+\mathrm{4} \\ $$

Commented by som(math1967) last updated on 17/Oct/20

Are you ex−student of Baranagar R.K.Mission

$$\mathrm{Are}\:\mathrm{you}\:\mathrm{ex}−\mathrm{student}\:\mathrm{of} \\ $$$$\mathrm{Baranagar}\:\mathrm{R}.\mathrm{K}.\mathrm{Mission} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

No sir. I am currently studying on Jadavpur Vidyapith

$${No}\:{sir}.\:{I}\:{am}\:{currently}\:{studying}\:{on}\:{Jadavpur}\:{Vidyapith} \\ $$

Commented by som(math1967) last updated on 17/Oct/20

Ok

$$\mathrm{Ok} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

Are you from Bengal sir?

$${Are}\:{you}\:{from}\:{Bengal}\:{sir}? \\ $$

Commented by som(math1967) last updated on 17/Oct/20

yes ,kolkata dunlop

$$\mathrm{yes}\:,\mathrm{kolkata}\:\mathrm{dunlop} \\ $$

Answered by som(math1967) last updated on 17/Oct/20

∫((sec^2 (x/2)dx)/(3sec^2 (x/2)−5×2sin(x/2)cos(x/2)sec^2 (x/2))) ∫((sec^2 (x/2)dx)/(3+3tan^2 (x/2)−10tan(x/2))) let tan(x/2)=z sec^2 (x/2)×(1/2)dx=dz sec^2 (x/2)dx=2dz 2∫(dz/(3z^2 −10z+3)) (2/3)∫(dz/(z^2 −((10z)/3)+1)) (2/3)∫(dz/(z^2 −2.z.(5/3)+((25)/9)+1−((25)/9))) (2/3)∫(dz/((z−(5/3))^2 −((4/3))^2 )) (2/3)×(3/(2×4 ))ln∣((z−(5/3)−(4/3))/(z−(5/3)+(4/3)))∣+c (1/4)ln∣((z−(9/3))/(z−(1/3)))∣+c ans z=tan(x/2)

$$\int\frac{\mathrm{sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx}}{\mathrm{3sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{5}×\mathrm{2sin}\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$\int\frac{\mathrm{sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx}}{\mathrm{3}+\mathrm{3tan}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{10tan}\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$\mathrm{let}\:\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{z} \\ $$$$\mathrm{sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{dx}=\mathrm{dz} \\ $$$$\mathrm{sec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx}=\mathrm{2dz} \\ $$$$\mathrm{2}\int\frac{\mathrm{dz}}{\mathrm{3z}^{\mathrm{2}} −\mathrm{10z}+\mathrm{3}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} −\frac{\mathrm{10z}}{\mathrm{3}}+\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} −\mathrm{2}.\mathrm{z}.\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{25}}{\mathrm{9}}+\mathrm{1}−\frac{\mathrm{25}}{\mathrm{9}}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{dz}}{\left(\mathrm{z}−\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{2}×\mathrm{4}\:}\mathrm{ln}\mid\frac{\mathrm{z}−\frac{\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}}{\mathrm{z}−\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{3}}}\mid+\mathrm{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\frac{\mathrm{z}−\frac{\mathrm{9}}{\mathrm{3}}}{\mathrm{z}−\frac{\mathrm{1}}{\mathrm{3}}}\mid+\mathrm{c}\:\:\mathrm{ans} \\ $$$$\mathrm{z}=\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}} \\ $$

Answered by 1549442205PVT last updated on 17/Oct/20

Put t=tan(x/2)⇒dt=(1/2)(1+t^2 )dx F=∫(( dt)/((t^2 +1)(3−((10t)/(1+t^2 )))))=∫((2dt)/((3t^2 −10t+3))) =(2/(3t^2 −10t+3))=(a/(3t−1))+(b/(t−3)) ⇔2=(a+3b)t−3a−b⇔ { ((3a+b=−2)),((a+3b=0)) :} ⇔b=1/4,a=−3/4 F=((−3)/4)∫(dt/(3t−1))+(1/4)∫(dt/(t−3))= (1/4)ln∣((t−3)/(3t−1))∣+C =(1/4)ln∣((tan(x/2)−3)/(3tan(x/2)−1))∣+C

$$\mathrm{Put}\:\mathrm{t}=\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{F}=\int\frac{\:\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}−\frac{\mathrm{10t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)}=\int\frac{\mathrm{2dt}}{\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{10t}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{10t}+\mathrm{3}}=\frac{\mathrm{a}}{\mathrm{3t}−\mathrm{1}}+\frac{\mathrm{b}}{\mathrm{t}−\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{2}=\left(\mathrm{a}+\mathrm{3b}\right)\mathrm{t}−\mathrm{3a}−\mathrm{b}\Leftrightarrow\begin{cases}{\mathrm{3a}+\mathrm{b}=−\mathrm{2}}\\{\mathrm{a}+\mathrm{3b}=\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{b}=\mathrm{1}/\mathrm{4},\mathrm{a}=−\mathrm{3}/\mathrm{4} \\ $$$$\mathrm{F}=\frac{−\mathrm{3}}{\mathrm{4}}\int\frac{\mathrm{dt}}{\mathrm{3t}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{dt}}{\mathrm{t}−\mathrm{3}}=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{3}}{\mathrm{3t}−\mathrm{1}}\mid+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\mathrm{ln}}\mid\frac{\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\mathrm{3}}{\mathrm{3}\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\mathrm{1}}\mid+\boldsymbol{\mathrm{C}} \\ $$

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