Question Number 109642 by malwan last updated on 24/Aug/20
Answered by Dwaipayan Shikari last updated on 24/Aug/20
$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{tan}^{\mathrm{2}} \theta+\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}{sec}^{\mathrm{2}} \theta{d}\theta\:\:\:\:\:\:\:\:\:\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}{tan}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\int{sec}\theta{d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{log}\left({sec}\theta+{tan}\theta\right)=\frac{\mathrm{1}}{\mathrm{5}}{log}\left(\sqrt{\frac{\mathrm{25}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}}\:+\frac{\mathrm{5}{x}}{\:\sqrt{\mathrm{2}}}\right)+{C} \\ $$
Commented by malwan last updated on 24/Aug/20
$${great}\:{Sir} \\ $$$${thank}\:{you}\:{so}\:{much} \\ $$