What is integral 1/(sinx+cosx) dx?

I tried solving it in many ways, but have no idea what to do to solve it.
I tried solving it in many ways, but have no idea what to do to solve it.
Given integral
\int \frac{1}{\sin x + \cos x} dx
Step 1: Multiply and divide the denominator with the root 2. Why because to convert the denominator in to the known formula.
\int \frac{1}{\frac{\sqrt{2}}{\sqrt{2}}(\sin x + \cos x)} dx
Step 2: Multiply denominator \sqrt{2} inside the bracket.
\int \frac{1}{\sqrt{2}(\sin x * \frac{1}{\sqrt{2}} + \cos x * \frac{1}{\sqrt{2}})} dx
Step 5: Find the value of \frac{1}{\sqrt{2}} in the form of angle (sin and cos) and substitute in the integral. because we are converting the equation into known formula sin(A + B) = sin A cos B + cos A sin B.
\int \frac{1}{\sqrt{2}(\sin x * \cos \frac{\pi}{4} + \cos x * \sin \frac{\pi}{4})} dx
Step 6: find this in integral sin A cos B + cos A sin B and replace it with sin(A + B)
(According to the question take the A and B values)
\int \frac{1}{\sqrt{2}}\frac{1}{\sin (x + \frac{\pi}{4})} dx
\int \frac{1}{\sqrt{2}} \cosec (x + \frac{\pi}{4}) dx (since sin x = \frac{1}{cosec x})
Step 7: Calculate it's value by substituting in the formula of the integration of cosec x
[NOTE] Using formula \int \cosec x dx = \ln ∣ \tan \frac{x}{2}∣ + C
or
\int \cosec x dx = ln |cosec x − cot x| = − ln |cosec x + cot x|
Step 8: For this question substitute the x = (x + \frac{\pi}{4}) value in the formulas mentioned in step 7, to get the answer.