Seems too easy! -2

First of all, note that

$\frac{d}{dx}(e^{g(x)}f(x)) = e^{g(x)} \{g'(x)f(x) + f'(x)\}$

Consequently,

$\int e^{g(x)} \{g'(x)f(x) + f'(x)\} dx = e^{g(x)}f(x)$

Our aim is to reduce the given integral to the above form.

We already have $g(x) = \sec(x)$

Let $I = \int_{0}^{\frac{\pi}{4}} e^{\sec x} {\frac{\sin (x+\frac{\pi}{4})}{(1-\sin x)(\cos x)}} dx$

Expanding $\sin (x+\frac{\pi}{4})$ ,

$I = \frac{1}{\sqrt2}\int_{0}^{\frac{\pi}{4}} e^{\sec x} {\frac{\sin x+\cos x}{(1-\sin x)(\cos x)}} dx$

Dividing Numerator and denominator by $\cos^2 x$ we have

$I = \frac{1}{\sqrt2}\int_{0}^{\frac{\pi}{4}} e^{\sec x} {\frac{\tan x\sec x+\sec x}{(\sec x-\tan x)}}dx$

$\Rightarrow I = \frac{1}{\sqrt2}\int_{0}^{\frac{\pi}{4}} e^{\sec x} \Big \{{\frac{\tan x\sec x}{(\sec x-\tan x)}} + \frac{\sec x}{(\sec x-\tan x)}\Big\}dx$

$\Rightarrow I = \frac{1}{\sqrt2}\int_{0}^{\frac{\pi}{4}} e^{\sec x} \Big \{\tan x\sec x\Big({\frac{1}{(\sec x-\tan x)}\Big)} + \frac{\sec x(\sec x-\tan x)}{(\sec x-\tan x)^2}\Big\}dx$

Comparing with the form above, $f(x) = \frac{1}{(\sec x-\tan x)}$

Thus, $I = \frac{1}{\sqrt2}\frac{e^{\sec x}}{(\sec x-\tan x)}\Big|_0^\frac{\pi}{4}$

Putting the limits, we have

$I = \frac{1}{\sqrt2}\Big( \frac{e^{\sqrt2}}{\sqrt2 - 1} - e\Big)$

$\Rightarrow I = \frac{(1+\sqrt2)e^{\sqrt2} - e}{\sqrt2}$

which gives

$a=1 , b=2 , c=2 , d=2$

$\Rightarrow a+b+c+d = 1+2+2+2 = \boxed{\color{#D61F06}7}$

In the given expression put t=sec(x) and then proceed.. Draw a right angled triangle with angle x and then put secx =t With this find all other trigonometric functions. Substitute these values in the given expression an spit into two terms . We will get second term as derivative of first term. Use the property and apply the limit. We will get a=1,b=2,c=2,d=2.

So,a+b+c+d=7.

DIDN'T PUT INTEGRAL SYMBOL PLEASE COPE WITH IT.

SORRY FOR PHOTO SOLUTION . I AM VERY SLOW WITH LATEX