Question

$e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

Answer 1

$e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

$e^x\tan ydx=-(1-e^x){\sec }^{2}ydy$

$\frac{e^x}{e^x-1}.dx=\frac{{\sec }^{2}y}{\tan y}.dy$

$\int \frac{e^x}{e^x-1}.dx=\int \frac{{\sec }^{2}y}{\tan y}.dy----\left(1\right)$

Put $e^x-1=u$ $\tan y=v$

$e^x.dx=du$ ${\sec }^{2}ydy=dv$

$e^x=\frac{du}{dx}$ ${\sec }^{2}ydy=dv$

$e^x=\frac{du}{dx}$ ${\sec }^{2}ydy=dv$

$dx=\frac{du}{e^x}$ $dy=\frac{dv}{{\sec }^{2}y}$

Put in $\left(1\right)$

$\int \frac{e^x}{u}.\frac{du}{e^x}=\int \frac{{\sec }^{2}y}{v}.\frac{dv}{{\sec }^{2}y}$

$\int \frac{du}{u}=\int \frac{dv}{v}$

$\log u+c_1=\log v$

$\log (e^x-1)+c_1=\log \left(\tan y\right)$

$\log |e^x-1|+\log c=\log |\tan y|$

$\log |c(e^x-1)|=\log |\tan y|$

$c(e^x-1)=\tan y$