Question Number 172420 by Physicien last updated on 26/Jun/22
Answered by thfchristopher last updated on 26/Jun/22
![∫(1/(7+tan x))dx Let t=tan (x/2) dt=(1/2)sec^2 xdx =(1/2)(1+t^2 )dc ⇒dx=((2dt)/(1+t^2 )) tan x=((2t)/(1−t^2 )) , sin x=((2t)/(1+t^2 )) , cos x=((1−t^2 )/(1+t^2 )) ∴ ∫(1/(7+tan x))dx =∫(((2dt)/(1+t^2 ))/(7+((2t)/(1−t^2 )))) =∫((2(1−t^2 )dt)/((1+t^2 )(7−7t^2 +2t))) ((2t^2 −2)/((t^2 +1)(7t^2 −2t−7)))=((At+B)/(t^2 +1))+((Ct+D)/(7t^2 −2t−7)) By solving, A=−(1/(25)), B=(7/(25)), C=(7/(25)), D=−(1/(25)) ((2t^2 −2)/((t^2 +1)(7t^2 −2t−7)))=(1/(25))(((7−t)/(t^2 +1))+((7t−1)/(7t^2 −2t−7))) ∫((2(1−t^2 )dt)/((1+t^2 )(7−7t^2 +2t))) =(1/(25))(∫((7−t)/(t^2 +1))dt+∫((7t−1)/(7t^2 −2t−7))dt) =(1/(50))[−∫((2t−14)/(t^2 +1))dt−∫((14t−2)/(7t^2 −2t−7))dt] =(1/(50))[14∫(dt/(t^2 +1))−ln ∣t^2 +1∣+ln ∣7t^2 −2t−7∣] =(7/(25))tan^(−1) t+(1/(50))ln ∣((7t^2 −2t−7)/(t^2 +1))∣+C =((7x)/(50))+(1/(50))ln ∣−sin x−7cos x∣+C =(1/(50))(7x+ln ∣sin x+7cos x∣)+C](Q172422.png)
$$\int\frac{\mathrm{1}}{\mathrm{7}+\mathrm{tan}\:{x}}{dx} \\ $$$$\mathrm{Let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$${dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \:{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dc} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:,\:\mathrm{sin}\:{x}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:,\:\mathrm{cos}\:{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\therefore\:\int\frac{\mathrm{1}}{\mathrm{7}+\mathrm{tan}\:{x}}{dx} \\ $$$$=\int\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{7}+\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\int\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{7}−\mathrm{7}{t}^{\mathrm{2}} +\mathrm{2}{t}\right)} \\ $$$$\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}\right)}=\frac{{At}+{B}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{Ct}+{D}}{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}} \\ $$$$\mathrm{By}\:\mathrm{solving}, \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{25}},\:{B}=\frac{\mathrm{7}}{\mathrm{25}},\:{C}=\frac{\mathrm{7}}{\mathrm{25}},\:{D}=−\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}\right)}=\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{\mathrm{7}−{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{7}{t}−\mathrm{1}}{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}}\right) \\ $$$$\int\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{7}−\mathrm{7}{t}^{\mathrm{2}} +\mathrm{2}{t}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\left(\int\frac{\mathrm{7}−{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\int\frac{\mathrm{7}{t}−\mathrm{1}}{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}}{dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\left[−\int\frac{\mathrm{2}{t}−\mathrm{14}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}−\int\frac{\mathrm{14}{t}−\mathrm{2}}{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\left[\mathrm{14}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\mid{t}^{\mathrm{2}} +\mathrm{1}\mid+\mathrm{ln}\:\mid\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}\mid\right] \\ $$$$=\frac{\mathrm{7}}{\mathrm{25}}\mathrm{tan}^{−\mathrm{1}} {t}+\frac{\mathrm{1}}{\mathrm{50}}\mathrm{ln}\:\mid\frac{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{7}}{{t}^{\mathrm{2}} +\mathrm{1}}\mid+{C} \\ $$$$=\frac{\mathrm{7}{x}}{\mathrm{50}}+\frac{\mathrm{1}}{\mathrm{50}}\mathrm{ln}\:\mid−\mathrm{sin}\:{x}−\mathrm{7cos}\:{x}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\left(\mathrm{7}{x}+\mathrm{ln}\:\mid\mathrm{sin}\:{x}+\mathrm{7cos}\:{x}\mid\right)+{C} \\ $$
Commented by Tawa11 last updated on 26/Jun/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by peter frank last updated on 03/Jul/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by MJS_new last updated on 26/Jun/22
![∫(dx/(7+tan x))= [t=tan x → dx=cos^2 x dt=(dt/(t^2 +1))] =∫(dt/((t+7)(t^2 +1)))= =(1/(50))∫(dt/(t+7))−(1/(100))∫((2t)/(t^2 +1))dt+(7/(50))∫(dt/(t^2 +1))= =(1/(50))ln (t+7) −(1/(100))ln (t^2 +1) +(7/(50))arctan t = =(1/(100))ln (((t+7)^2 )/(t^2 +1)) +(7/(50))arctan t = =(7/(50))x+(1/(50))ln ∣7cos x +sin x∣ +C](Q172424.png)
$$\int\frac{{dx}}{\mathrm{7}+\mathrm{tan}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{{dt}}{\left({t}+\mathrm{7}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\int\frac{{dt}}{{t}+\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{100}}\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\frac{\mathrm{7}}{\mathrm{50}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\mathrm{ln}\:\left({t}+\mathrm{7}\right)\:−\frac{\mathrm{1}}{\mathrm{100}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:+\frac{\mathrm{7}}{\mathrm{50}}\mathrm{arctan}\:{t}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{100}}\mathrm{ln}\:\frac{\left({t}+\mathrm{7}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{7}}{\mathrm{50}}\mathrm{arctan}\:{t}\:= \\ $$$$=\frac{\mathrm{7}}{\mathrm{50}}{x}+\frac{\mathrm{1}}{\mathrm{50}}\mathrm{ln}\:\mid\mathrm{7cos}\:{x}\:+\mathrm{sin}\:{x}\mid\:+{C} \\ $$
Commented by thfchristopher last updated on 26/Jun/22
$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{simple}\:\mathrm{than}\:\mathrm{mine}.\:\mathrm{Thanks}! \\ $$
Commented by MJS_new last updated on 26/Jun/22
$$\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{because} \\ $$$${t}=\mathrm{tan}\:{x}\:\mathrm{works}\:\mathrm{too}.\:\mathrm{To}\:\mathrm{see}\:\mathrm{this}\:\mathrm{needs}\:\mathrm{just}\:\mathrm{a} \\ $$$$\mathrm{little}\:\mathrm{experience}. \\ $$
Commented by Tawa11 last updated on 26/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$