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