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Question Number 30258 by mondodotto@gmail.com last updated on 19/Feb/18

Answered by $@ty@m last updated on 19/Feb/18

Let sec x−1=z tan^2 xdx=dz dx=(dz/(sec^2 x−1))=(dz/(z(z+2))) I=∫(dx/(z^2 (z+2))) Let (1/(z^2 (z+2)))=(A/z^2 )+(B/(z+2))+(C/z) ⇒A(z+2)+Bz^2 +Cz(z+2)≡1 ⇒A=(1/2), B=(1/4)& A+B−C=1⇒C=((−1)/4) ∴I=∫(dz/(2z^2 ))+∫(dz/(4(z+2)))−∫(dz/(4z)) =(1/2)(((−1)/z))+(1/4)ln (z+2)−(1/4)ln z+C = ((−1)/(2(sec x−1)))+(1/4)ln (sec x+1)−(1/4)ln (sec x−1)+C

$${Let}\:\mathrm{sec}\:{x}−\mathrm{1}={z} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {xdx}={dz} \\ $$$${dx}=\frac{{dz}}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}=\frac{{dz}}{{z}\left({z}+\mathrm{2}\right)} \\ $$$${I}=\int\frac{{dx}}{{z}^{\mathrm{2}} \left({z}+\mathrm{2}\right)}\: \\ $$$${Let}\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \left({z}+\mathrm{2}\right)}=\frac{{A}}{{z}^{\mathrm{2}} }+\frac{{B}}{{z}+\mathrm{2}}+\frac{{C}}{{z}} \\ $$$$\Rightarrow{A}\left({z}+\mathrm{2}\right)+{Bz}^{\mathrm{2}} +{Cz}\left({z}+\mathrm{2}\right)\equiv\mathrm{1} \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}},\:{B}=\frac{\mathrm{1}}{\mathrm{4}}\& \\ $$$${A}+{B}−{C}=\mathrm{1}\Rightarrow{C}=\frac{−\mathrm{1}}{\mathrm{4}} \\ $$$$\therefore{I}=\int\frac{{dz}}{\mathrm{2}{z}^{\mathrm{2}} }+\int\frac{{dz}}{\mathrm{4}\left({z}+\mathrm{2}\right)}−\int\frac{{dz}}{\mathrm{4}{z}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{1}}{{z}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({z}+\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{z}+{C} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{sec}\:{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{sec}\:{x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{sec}\:{x}−\mathrm{1}\right)+{C} \\ $$

Commented by rahul 19 last updated on 19/Feb/18

try my ques. then :)

$$\left.\mathrm{try}\:\mathrm{my}\:\mathrm{ques}.\:\mathrm{then}\::\right) \\ $$

Commented by $@ty@m last updated on 19/Feb/18

Not as challanging as claimed. :(

$${Not}\:{as}\:{challanging}\:{as}\:{claimed}. \\ $$$$:\left(\right. \\ $$

Answered by ajfour last updated on 19/Feb/18

∫(dx/(sec x−1)) = ∫((sec x+1)/(sec^2 x−1))dx =∫ (((sec x+1))/(tan^2 x))dx =∫cosec xcot xdx +∫cot^2 xdx =−cosec x+c_1 +∫(cosec^2 x−1)dx = −cosec x−cot x−x−c .

$$\int\frac{{dx}}{\mathrm{sec}\:{x}−\mathrm{1}}\:=\:\int\frac{\mathrm{sec}\:{x}+\mathrm{1}}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}{dx} \\ $$$$=\int\:\frac{\left(\mathrm{sec}\:{x}+\mathrm{1}\right)}{\mathrm{tan}\:^{\mathrm{2}} {x}}{dx}\: \\ $$$$=\int\mathrm{cosec}\:{x}\mathrm{cot}\:{xdx}\:+\int\mathrm{cot}\:^{\mathrm{2}} {xdx} \\ $$$$=−\mathrm{cosec}\:{x}+{c}_{\mathrm{1}} +\int\left(\mathrm{cosec}\:^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$\:=\:−\mathrm{cosec}\:{x}−\mathrm{cot}\:{x}−{x}−{c}\:. \\ $$

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