$\int e^{\sin x}.(xCosx-\tan x.Secx)dx=$

$\int e^{\sin x}.(xCosx-\tan x.Secx)dx$

$$\begin{gathered} =\int e^{\sin x}\left[\left(x-secx\right)\cos x+1-\tan xsecx\right]dx \\ \left(\therefore secx\cos x=1\right) \\ =\int e^{\sin x}\left[\cos x\left(x-secx\right)+\left(1-secx\tan x\right)\right]dx \\ =\int e^{g\left(x\right)}\left[g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)\right]dx \\ =\int e^{g\left(x\right)}\left[g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)\right]dx where f\left(x\right)=x-secx \\ g\left(x\right)=\sin x \\ =e^{g\left(x\right)}.f\left(x\right)+c \\ =e^{\sin x}\left(x-secx\right)+c \end{gathered}$$