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1-1-cot-x-dx-

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Question Number 5838 by gourav~ last updated on 31/May/16
∫(1/(1+cot x))dx
$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cot}\:{x}}{dx} \\ $$
Commented by Yozzii last updated on 31/May/16
(1/(1+cotx))=((sinx)/(cosx+sinx)) Let t=tan0.5x⇒dt=0.5sec^2 0.5dx dx=(2/(1+t^2 ))dt (1+tan^2 0.5x=sec^2 .5x) cosx=((1−t^2 )/(1+t^2 )), sinx=((2t)/(1+t^2 )) ∴∫(1/(1+cotx))dx=∫((2t×2)/(1−t^2 +2t))dt =−2∫((−2+2−2t)/(1+2t−t^2 ))dt =−2{∫(((d/dt)(1+2t−t^2 ))/(1+2t−t^2 ))dt−2∫(dt/(2−(t−1)^2 ))} ∫(dx/(1+cotx))=−2ln∣1+2t−t^2 ∣+4∫(dt/(((√2)−t+1)((√2)+t−1))) (1/( (√2)+1−t))+(1/( (√2)−1+t))=(((√2)−1+t+(√2)+1−t)/(2−(t−1)^2 )) =((2(√2))/(2−(t−1)^2 )) ⇒(1/(2(√2)))((1/(t+(√2)−1))+(1/( (√2)+1−t)))=(1/(((√2)−1+t)((√2)+1−t))) ∴∫(dx/(1+cotx))=(√2)∫((1/(t+(√2)−1))−(1/(−(√2)−1+t)))dt+ln((1/((1+2t−t^2 )^2 ))) ∫(dx/(1+cotx))=(√2)ln∣((t−1+(√2))/(t−1−(√2)))∣+ln((1/((1+2t−t^2 )^2 )))+C ∫(dx/(1+cotx))=ln{((∣tan0.5x−1+(√2)∣^(√2) )/(∣tan0.5x−1−(√2)∣^(√2) (1+2tan0.5x−tan^2 0.5x)^2 ))}+C ∫(dx/(1+cotx))=(√2)ln∣t−1+(√2)∣−(√2)ln∣t−1−(√2)∣−2ln∣t−1+(√2)∣−2ln∣t−1−(√2)∣+C ∫(dx/(1+cotx))=((√2)−2)ln∣t−1+(√2)∣−((√2)+2)ln∣t−1−(√2)∣+C ∫(dx/(1+cotx))=ln∣(1/(∣tan0.5x−1+(√2)∣^(2−(√2)) ∣tan0.5x−1−(√2)∣^(2+(√2)) ))∣+C
$$\frac{\mathrm{1}}{\mathrm{1}+{cotx}}=\frac{{sinx}}{{cosx}+{sinx}} \\ $$$${Let}\:{t}={tan}\mathrm{0}.\mathrm{5}{x}\Rightarrow{dt}=\mathrm{0}.\mathrm{5}{sec}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{dx} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{x}={sec}^{\mathrm{2}} .\mathrm{5}{x}\right) \\ $$$${cosx}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} },\:{sinx}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\therefore\int\frac{\mathrm{1}}{\mathrm{1}+{cotx}}{dx}=\int\frac{\mathrm{2}{t}×\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}}{dt} \\ $$$$=−\mathrm{2}\int\frac{−\mathrm{2}+\mathrm{2}−\mathrm{2}{t}}{\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} }{dt} \\ $$$$=−\mathrm{2}\left\{\int\frac{\frac{{d}}{{dt}}\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} }{dt}−\mathrm{2}\int\frac{{dt}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}=−\mathrm{2}{ln}\mid\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \mid+\mathrm{4}\int\frac{{dt}}{\left(\sqrt{\mathrm{2}}−{t}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}+{t}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}−{t}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}+{t}}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}+{t}+\sqrt{\mathrm{2}}+\mathrm{1}−{t}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{{t}+\sqrt{\mathrm{2}}−\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}−{t}}\right)=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}−\mathrm{1}+{t}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}−{t}\right)} \\ $$$$\therefore\int\frac{{dx}}{\mathrm{1}+{cotx}}=\sqrt{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{t}+\sqrt{\mathrm{2}}−\mathrm{1}}−\frac{\mathrm{1}}{−\sqrt{\mathrm{2}}−\mathrm{1}+{t}}\right){dt}+{ln}\left(\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right) \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}=\sqrt{\mathrm{2}}{ln}\mid\frac{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}\mid+{ln}\left(\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)+{C} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}={ln}\left\{\frac{\mid{tan}\mathrm{0}.\mathrm{5}{x}−\mathrm{1}+\sqrt{\mathrm{2}}\mid^{\sqrt{\mathrm{2}}} }{\mid{tan}\mathrm{0}.\mathrm{5}{x}−\mathrm{1}−\sqrt{\mathrm{2}}\mid^{\sqrt{\mathrm{2}}} \left(\mathrm{1}+\mathrm{2}{tan}\mathrm{0}.\mathrm{5}{x}−{tan}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{x}\right)^{\mathrm{2}} }\right\}+{C} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}=\sqrt{\mathrm{2}}{ln}\mid{t}−\mathrm{1}+\sqrt{\mathrm{2}}\mid−\sqrt{\mathrm{2}}{ln}\mid{t}−\mathrm{1}−\sqrt{\mathrm{2}}\mid−\mathrm{2}{ln}\mid{t}−\mathrm{1}+\sqrt{\mathrm{2}}\mid−\mathrm{2}{ln}\mid{t}−\mathrm{1}−\sqrt{\mathrm{2}}\mid+{C} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}=\left(\sqrt{\mathrm{2}}−\mathrm{2}\right){ln}\mid{t}−\mathrm{1}+\sqrt{\mathrm{2}}\mid−\left(\sqrt{\mathrm{2}}+\mathrm{2}\right){ln}\mid{t}−\mathrm{1}−\sqrt{\mathrm{2}}\mid+{C} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}={ln}\mid\frac{\mathrm{1}}{\mid{tan}\mathrm{0}.\mathrm{5}{x}−\mathrm{1}+\sqrt{\mathrm{2}}\mid^{\mathrm{2}−\sqrt{\mathrm{2}}} \mid{tan}\mathrm{0}.\mathrm{5}{x}−\mathrm{1}−\sqrt{\mathrm{2}}\mid^{\mathrm{2}+\sqrt{\mathrm{2}}} }\mid+{C} \\ $$$$ \\ $$
Commented by Yozzii last updated on 31/May/16
Let R=∫((sinx)/(cosx+sinx))dx and J=∫((cosx)/(cosx+sinx))dx. ∴R+J=∫1dx=x+C J−R=∫((cosx−sinx)/(cosx+sinx))dx=∫(((d/dx)(cosx+sinx))/(cosx+sinx))dx J−R=ln∣cosx+sinx∣+N ∴R+J−J+R=x−ln∣cosx+sinx∣+C−N R=0.5(x−ln∣cosx+sinx∣)+P ∫(dx/(1+cotx))=(1/2)(x−ln∣cosx+sinx∣)+P P=constant of integration.
$${Let}\:{R}=\int\frac{{sinx}}{{cosx}+{sinx}}{dx}\:{and}\:{J}=\int\frac{{cosx}}{{cosx}+{sinx}}{dx}. \\ $$$$\therefore{R}+{J}=\int\mathrm{1}{dx}={x}+{C} \\ $$$${J}−{R}=\int\frac{{cosx}−{sinx}}{{cosx}+{sinx}}{dx}=\int\frac{\frac{{d}}{{dx}}\left({cosx}+{sinx}\right)}{{cosx}+{sinx}}{dx} \\ $$$${J}−{R}={ln}\mid{cosx}+{sinx}\mid+{N} \\ $$$$\therefore{R}+{J}−{J}+{R}={x}−{ln}\mid{cosx}+{sinx}\mid+{C}−{N} \\ $$$${R}=\mathrm{0}.\mathrm{5}\left({x}−{ln}\mid{cosx}+{sinx}\mid\right)+{P} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{cotx}}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{ln}\mid{cosx}+{sinx}\mid\right)+{P} \\ $$$${P}={constant}\:{of}\:{integration}. \\ $$$$ \\ $$$$ \\ $$
Commented by gourav~ last updated on 01/Jun/16
thank you.=.
$${thank}\:{you}.=. \\ $$

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