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Question Number 12766 by Joel577 last updated on 01/May/17
∫ (dx/(1 + tan x))
$$\int\:\frac{{dx}}{\mathrm{1}\:+\:\mathrm{tan}\:{x}} \\ $$
Answered by sma3l2996 last updated on 01/May/17
t=tanx⇒dt=(1+tan^2 x)dx dx=(dt/(1+t^2 )) ∫(dx/(1+tanx))=∫(dt/((1+t)(1+t^2 ))) (1/((1+t)(1+t^2 )))=(a/(1+t))+((bt+c)/(1+t^2 )) a=(1/2) ; c=(1/2) ; b=((−1)/2) ∫(dt/((1+t)(1+t^2 )))=(1/2)∫(dt/(1+t))−(1/2)∫((t−1)/(1+t^2 ))dt =(1/2)ln∣1+t∣−(1/4)∫((2t)/(1+t^2 ))dt+(1/2)∫(dt/(1+t^2 ))+c =(1/2)ln∣1+t∣−(1/4)ln∣1+t^2 ∣+(1/2)tan^(−1) (t)+C =(1/2)(ln∣((1+t)/( (√(1+t^2 ))))∣+tan^(−1) (t))+C =(1/2)(ln∣((1+tanx)/( (√(1+tan^2 x))))∣+x)+C =(1/2)(ln∣cosx(1+tanx)∣+x)+C
$${t}={tanx}\Rightarrow{dt}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$${dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dx}}{\mathrm{1}+{tanx}}=\int\frac{{dt}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{{a}}{\mathrm{1}+{t}}+\frac{{bt}+{c}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:{b}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\int\frac{{dt}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}+{t}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{t}\mid−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{t}\mid−\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\mathrm{1}+{t}^{\mathrm{2}} \mid+\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\mid\frac{\mathrm{1}+{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\mid+{tan}^{−\mathrm{1}} \left({t}\right)\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\mid\frac{\mathrm{1}+{tanx}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}\mid+{x}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\mid{cosx}\left(\mathrm{1}+{tanx}\right)\mid+{x}\right)+{C} \\ $$$$ \\ $$
Commented by Joel577 last updated on 01/May/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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