$$\int_{0}^{\frac{\pi}{4}}\frac{e^{\sec x}\cdot \sin \bigg(x+\frac{\pi}{4}\bigg)}{(1-\sin x)\cdot \cos x}dx$$
Try: substitute $\displaystyle x+\frac{\pi}{4}=t$ then $dx=dt$
So $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{\frac{1}{\cos(t-45^\circ)}}\cdot \frac{\sin t}{1-\sin(t-45^\circ)\cdot \cos(t-45^\circ)}dx$$
Could some help me to solve it , thanks
Hint:
$$e^{\sec x}\dfrac{\sqrt2\sin\left(\dfrac\pi4+x\right)}{(1-\sin x)\cos x}$$
$$=e^{\sec x}\dfrac{(1+\sin x)(\sin x+\cos x)}{\cos^3x}$$
$$=e^{\sec x}(\sec^2x+\sec x\tan x)(1+\tan x)$$
$$=e^{\sec x}\sec^2x+\dfrac{d(e^{\sec x})}{dx}\sec x+\dfrac{d(e^{\sec x})}{dx}+\dfrac{d(e^{\sec x})}{dx}\tan x$$
Now integrate by parts,
$$\int e^{\sec x}\sec^2xdx= \int e^{\sec x}(d\tan x)=?$$
$$\int\dfrac{d(e^{\sec x})}{dx}\sec x\ dx=\sec x\cdot d(e^{\sec x})=?$$
$$\dfrac{d(e^{\sec x})}{dx}\tan x=\tan x\cdot d(e^{\sec x})=?$$