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Question Number 120758 by bramlexs22 last updated on 02/Nov/20

∫ (dx/(a sin x + b cos x))

$$\:\int\:\frac{\mathrm{dx}}{{a}\:\mathrm{sin}\:\mathrm{x}\:+\:{b}\:\mathrm{cos}\:\mathrm{x}} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Nov/20

2∫(dt/(a(((2t)/(1+t^2 )))+b(((1−t^2 )/(1+t^2 ))))).(1/(1+t^2 )) tan(x/2)=t =2∫(dt/(2at+b−bt^2 )) =−(2/b)∫(dt/(t^2 −((2at)/b)−1)) =−(2/b)∫(dt/((t−(a/b))^2 −(a^2 /b^2 )−1)) =(2/b)∫(dt/( ((√((a^2 /b^2 )+1)))^2 −(t−(a/b))^2 )) =(2/b).(1/(2(√((a^2 /b^2 )+1))))log(((t−(a/b)+(√((a^2 /b^2 )+1)))/(t−(a/b)−(√((a^2 /b^2 )+1)))))+C =(1/(b(√((a^2 /b^2 )+1)))).log(((tan(x/2)−(a/b)+(√((a^2 /b^2 )+1)))/(tan(x/2)−(a/b)−(√((a^2 /b^2 )+1)))))+C =(1/( (√(a^2 +b^2 ))))log(((btan(x/2)−a+(√(a^2 +b^2 )))/(btan(x/2)−a−(√(a^2 +b^2 )))))+C

$$\mathrm{2}\int\frac{{dt}}{{a}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)+{b}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\mathrm{2}{at}+{b}−{bt}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{2}}{{b}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\frac{\mathrm{2}{at}}{{b}}−\mathrm{1}} \\ $$$$=−\frac{\mathrm{2}}{{b}}\int\frac{{dt}}{\left({t}−\frac{{a}}{{b}}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{{b}}\int\frac{{dt}}{\:\left(\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}\right)^{\mathrm{2}} −\left({t}−\frac{{a}}{{b}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{{b}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}{log}\left(\frac{{t}−\frac{{a}}{{b}}+\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}{{t}−\frac{{a}}{{b}}−\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{{b}\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}.{log}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}−\frac{{a}}{{b}}+\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}{{tan}\frac{{x}}{\mathrm{2}}−\frac{{a}}{{b}}−\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{log}\left(\frac{{btan}\frac{{x}}{\mathrm{2}}−{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{btan}\frac{{x}}{\mathrm{2}}−{a}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)+{C} \\ $$

Answered by liberty last updated on 02/Nov/20

find the two numbers r and β such that a = r cos β , b = r sin β → { ((r =(√(a^2 +b^2 )))),((β=tan^(−1) ((b/a)))) :} ∴ ∫ (dx/(a sin x + b cos x)) = (1/r)∫ (dx/(sin (x+β))) =(1/r) ∫ cosec (x+β) dx = (1/r) ln (tan (((x+β)/2)))+c = (1/( (√(a^2 +b^2 )))) ln (tan (((x+β)/2))) + c =(1/( (√(a^2 +b^2 )))) ln (tan (((tan^(−1) ((b/a))+x)/2))) +c

$$\mathrm{find}\:\mathrm{the}\:\mathrm{two}\:\mathrm{numbers}\:\mathrm{r}\:\mathrm{and}\:\beta\:\mathrm{such}\:\mathrm{that}\: \\ $$$${a}\:=\:{r}\:\mathrm{cos}\:\beta\:,\:{b}\:=\:{r}\:\mathrm{sin}\:\beta\:\rightarrow\begin{cases}{{r}\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\\{\beta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}\end{cases} \\ $$$$\therefore\:\int\:\frac{\mathrm{dx}}{{a}\:\mathrm{sin}\:{x}\:+\:{b}\:\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{{r}}\int\:\frac{\mathrm{dx}}{\mathrm{sin}\:\left(\mathrm{x}+\beta\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{r}}\:\int\:\mathrm{cosec}\:\left(\mathrm{x}+\beta\right)\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{r}}\:\mathrm{ln}\:\left(\mathrm{tan}\:\left(\frac{\mathrm{x}+\beta}{\mathrm{2}}\right)\right)+\mathrm{c}\: \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{ln}\:\left(\mathrm{tan}\:\left(\frac{{x}+\beta}{\mathrm{2}}\right)\right)\:+\:\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{ln}\:\left(\mathrm{tan}\:\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)+{x}}{\mathrm{2}}\right)\right)\:+{c}\: \\ $$

Answered by 675480065 last updated on 02/Nov/20

asinx+bcosx=a(((2t)/(1+t^2 )))+b(((1−t^2 )/(1+t^2 ))) t=tan((x/2))⇒2dt=sec^2 ((x/2))dx ⇒dx=((2dt)/(1+t^2 )) I = ∫((2dt)/((1+t^2 ))).(((1+t^2 ))/(2at+b(1−t^2 ))) = 2∫(dt/((2at−bt^2 +b))) I = −2∫(dt/((bt^2 −2at−b))) =−2∫(dt/(b(t^2 −2at−b+(a)^2 −(a^2 )))) ⇒ I = −2∫(dt/(b[(t−a)^2 −(a^2 +b)])) ⇒ I = −(2/b)∫(dt/((t−a)^2 −(√((a^2 +b)^2 )))) Let (t−a)=(√(a^2 +b)) tanhβ ⇒dt = (√(a^2 +b)) sech^2 βdβ ⇒ I = −(2/b)∫(((√(a^2 +b)) sech^2 βdβ)/( (√((a^2 +b)^2 ))(tanh^2 β−1))) ⇒ I = −(2/(b(√(a^2 +b))))∫dβ = −2((1/(b(√(a^2 +b)))))tanh^(−1) (((t−a)/( (√(a^2 +b))))) + k ⇒ I = −(1/(b(√(a^2 +b))))ln∣((t−a+(√(a^2 +b)))/(t−a−(√(a^2 +b))))∣ + k

$${asinx}+{bcosx}={a}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)+{b}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$${t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\Rightarrow\mathrm{2}{dt}={sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}\:=\:\int\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}.\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{at}+{b}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\:\mathrm{2}\int\frac{{dt}}{\left(\mathrm{2}{at}−{bt}^{\mathrm{2}} +{b}\right)} \\ $$$${I}\:=\:−\mathrm{2}\int\frac{{dt}}{\left({bt}^{\mathrm{2}} −\mathrm{2}{at}−{b}\right)}\:=−\mathrm{2}\int\frac{{dt}}{{b}\left({t}^{\mathrm{2}} −\mathrm{2}{at}−{b}+\left({a}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} \right)\right)} \\ $$$$\Rightarrow\:{I}\:=\:−\mathrm{2}\int\frac{{dt}}{{b}\left[\left({t}−{a}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}\right)\right]} \\ $$$$\Rightarrow\:{I}\:=\:−\frac{\mathrm{2}}{{b}}\int\frac{{dt}}{\left({t}−{a}\right)^{\mathrm{2}} −\sqrt{\left({a}^{\mathrm{2}} +{b}\right)^{\mathrm{2}} }} \\ $$$${Let}\:\left({t}−{a}\right)=\sqrt{{a}^{\mathrm{2}} +{b}}\:{tanh}\beta \\ $$$$\Rightarrow{dt}\:=\:\sqrt{{a}^{\mathrm{2}} +{b}}\:{sech}^{\mathrm{2}} \beta{d}\beta \\ $$$$\Rightarrow\:{I}\:=\:−\frac{\mathrm{2}}{{b}}\int\frac{\sqrt{{a}^{\mathrm{2}} +{b}}\:{sech}^{\mathrm{2}} \beta{d}\beta}{\:\sqrt{\left({a}^{\mathrm{2}} +{b}\right)^{\mathrm{2}} }\left({tanh}^{\mathrm{2}} \beta−\mathrm{1}\right)} \\ $$$$\Rightarrow\:{I}\:=\:−\frac{\mathrm{2}}{{b}\sqrt{{a}^{\mathrm{2}} +{b}}}\int{d}\beta\:=\:−\mathrm{2}\left(\frac{\mathrm{1}}{{b}\sqrt{{a}^{\mathrm{2}} +{b}}}\right){tanh}^{−\mathrm{1}} \left(\frac{{t}−{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}}}\right)\:+\:{k} \\ $$$$\Rightarrow\:{I}\:=\:−\frac{\mathrm{1}}{{b}\sqrt{{a}^{\mathrm{2}} +{b}}}{ln}\mid\frac{\mathrm{t}−{a}+\sqrt{{a}^{\mathrm{2}} +{b}}}{{t}−{a}−\sqrt{{a}^{\mathrm{2}} +{b}}}\mid\:+\:\mathrm{k} \\ $$

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