Question Number 149996 by tabata last updated on 08/Aug/21
$$\left(\mathrm{1}\right)\:\int\:\:\frac{{dx}}{\mathrm{1}+{tanx}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int\:\:\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$
Answered by mindispower last updated on 08/Aug/21
$$=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\sqrt{{tg}\left({x}\right)}}{{tg}\left({x}\right)}{dx}=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\mathrm{1}}{\:\sqrt{{tg}\left({x}\right)}}{dx} \\ $$$$=\mathrm{2}\int\frac{{dtg}\left({x}\right)}{\mathrm{2}\sqrt{{tg}\left({x}\right)}}=\mathrm{2}\sqrt{{tg}\left({x}\right)}+{c} \\ $$
Commented by tabata last updated on 08/Aug/21
$${thank}\:{you}\:{alot}\:{sir} \\ $$
Answered by mindispower last updated on 08/Aug/21
$$\left(\mathrm{1}\right)=\int\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{cos}\left({x}\right)+{sin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}+\int\frac{{cos}\left({x}\right)−{sin}\left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cos}\left({x}\right)+{sin}\left({x}\right)\mid+{c} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{{ln}\mid\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{2}}\right)\mid}{\mathrm{2}}+{c} \\ $$
Commented by puissant last updated on 08/Aug/21
$${nice}\:{sir}\:{mindispower}… \\ $$
Commented by tabata last updated on 08/Aug/21
$${thank}\:{you}\:{sir} \\ $$
Answered by puissant last updated on 08/Aug/21
$$\left.\mathrm{2}\right) \\ $$$${Q}=\int\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$$${u}=\sqrt{{tanx}}\rightarrow{u}^{\mathrm{2}} ={tanx}\rightarrow\mathrm{2}{udu}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\rightarrow\:{dx}=\mathrm{2}{ucos}^{\mathrm{2}} {xdu} \\ $$$$\Rightarrow\:{Q}=\int\frac{{u}}{{sinx}\:{cosx}}\mathrm{2}{ucos}^{\mathrm{2}} {xdu} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\int{u}^{\mathrm{2}} {cotanxdu} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\int{u}^{\mathrm{2}} \frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du}\:=\:\mathrm{2}\int{du}\:=\:\mathrm{2}{u}+{C}.. \\ $$$$\:\:\:\:\because\:{Q}\:=\:\mathrm{2}\sqrt{{tanx}}+{C}.. \\ $$$$ \\ $$