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Question Number 28706 by students last updated on 29/Jan/18

most important question gor boar or iit solve the integration (1/(3sinx+4cosx))

$${most}\:{important}\:{question}\:{gor}\:{boar}\:{or}\:{iit}\: \\ $$$${solve}\:{the}\:{integration}\:\:\frac{\mathrm{1}}{\mathrm{3}{sinx}+\mathrm{4}{cosx}} \\ $$

Commented by abdo imad last updated on 29/Jan/18

let put I= ∫ (dx/(3sinx+4cosx)) and use the ch. tan((x/2))=t I= ∫ (1/(((6t)/(1+t^2 )) +4((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 ))= ∫ ((2dt)/(6t +4−4t^2 )) = ∫ (dt/(−2t^2 +3t +2))=−∫ (dt/(2t^2 −3t−2)) let find the roots of F(t)= (1/(2t^2 −3t−2)) Δ= 9 −4(−4)=25⇒t_1 =((3+5)/4)=2 t_2 =((3−5)/4)=((−1)/2) so F(t)= (1/(2(t−2)(t+(1/2))))=(a/(t−2)) +(b/(t+(1/2))) a=lim_(t→2) (t−2)F(t)= (1/(2(2+(1/2))))= (1/5) b=lim_(t→−(1/2)) (t+(1/2))F(t)= (1/(2(−(1/2)−2)))=−(1/5) F(t)=(1/5)( (1/(t−2)) − (1/(t+(1/2)))) ∫F(t)dt= (1/5)ln∣((t−2)/(t+(1/2)))∣+λ and I=−(1/5)ln∣((t−2)/(t+(1/2)))∣ +λ= −(1/5)ln∣((tan((x/2))−2)/(tan((x/2))+(1/2)))∣ +λ

$${let}\:{put}\:{I}=\:\int\:\:\frac{{dx}}{\mathrm{3}{sinx}+\mathrm{4}{cosx}}\:{and}\:{use}\:{the}\:{ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${I}=\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{6}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{4}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\:\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{6}{t}\:+\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\frac{{dt}}{−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}{t}\:+\mathrm{2}}=−\int\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}}\:{let}\:{find}\:{the}\:{roots}\:{of} \\ $$$${F}\left({t}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}}\:\:\:\Delta=\:\mathrm{9}\:−\mathrm{4}\left(−\mathrm{4}\right)=\mathrm{25}\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{4}}=\mathrm{2} \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{4}}=\frac{−\mathrm{1}}{\mathrm{2}}\:{so}\:{F}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{2}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{{a}}{{t}−\mathrm{2}}\:+\frac{{b}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${a}={lim}_{{t}\rightarrow\mathrm{2}} \left({t}−\mathrm{2}\right){F}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${b}={lim}_{{t}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right){F}\left({t}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)}=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{5}}\left(\:\:\frac{\mathrm{1}}{{t}−\mathrm{2}}\:−\:\frac{\mathrm{1}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$\int{F}\left({t}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid\frac{{t}−\mathrm{2}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}}\mid+\lambda\:\:\:\:\:{and} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid\frac{{t}−\mathrm{2}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}}\mid\:+\lambda=\:−\frac{\mathrm{1}}{\mathrm{5}}{ln}\mid\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}\mid\:+\lambda \\ $$

Commented by students last updated on 29/Jan/18

sir i am confused please sir solve the question other method

$${sir}\:{i}\:{am}\:{confused}\:{please}\:{sir}\:{solve}\:{the}\:{question}\:{other}\:{method} \\ $$

Commented by Tinkutara last updated on 29/Jan/18

This is a similar question. Just ignore the limits and change the values.

Commented by Tinkutara last updated on 29/Jan/18

Commented by Tinkutara last updated on 29/Jan/18

Commented by students last updated on 29/Jan/18

thank u sir

$${thank}\:{u}\:{sir} \\ $$

Commented by abdo imad last updated on 29/Jan/18

i have another method with residus theorem but its more difficult than this and its better for you to look the solution given by sir tinkutara it s the same without borns...

$${i}\:{have}\:{another}\:{method}\:{with}\:{residus}\:{theorem}\:{but}\:{its} \\ $$$${more}\:{difficult}\:{than}\:{this}\:{and}\:{its}\:{better}\:{for}\:{you}\:{to}\:{look}\:{the} \\ $$$${solution}\:{given}\:{by}\:{sir}\:{tinkutara}\:\:{it}\:{s}\:{the}\:{same}\:{without} \\ $$$${borns}... \\ $$

Answered by ajfour last updated on 29/Jan/18

∫(dx/(3sin x+4cos x))=∫(dx/(5sin (x+α))) where cos α=(3/5), sin α=(4/5) =(1/5)∫cosec (x+α)dx =(1/5)ln ∣cosec (x+α)−cot (x+α)∣+c =(1/5)ln ∣((5−3cos x+4sin x)/(3sin x+4cos x))∣+c

$$\int\frac{{dx}}{\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}}=\int\frac{{dx}}{\mathrm{5sin}\:\left({x}+\alpha\right)} \\ $$$${where}\:\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}},\:\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int\mathrm{cosec}\:\left({x}+\alpha\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid\mathrm{cosec}\:\left({x}+\alpha\right)−\mathrm{cot}\:\left({x}+\alpha\right)\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid\frac{\mathrm{5}−\mathrm{3cos}\:{x}+\mathrm{4sin}\:{x}}{\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}}\mid+{c}\: \\ $$

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