Challenging Integration involving cosecant and cosine The question is
$\int \frac{cosec^2x -2020}{cos^{2020}x}$=$\frac{Af(x)^B}{g(x)^{2020}}$+c where f($\frac{\pi}{6}$)=$\sqrt3$
Then, the value of $A^2+B^2+f(-\frac{\pi}{4}$)=....
I tried converting into tangent form , but it's not making sense further, I tried by by parts formula,  but that also makes it quite long, I am not getting any ideas I am now trying this for much time.
It must be of some specific form , but it is not hiting.
I would like to have some hints to carry on...it's a high school level question, so I expect it to be easy, once it strikes. :-)
 A: The "form" of the solution is already provided to you. Upon analysis, it seems to me that the first step should be to multiply both the numerator and denominator by $cos^{2020} x$.
Why? because, the derivative of the expression given should be the integrand. Recall the quotient rule, which says that the denominator should be squared. So, having $cos^{4040}x$ draws us one step closer to transforming the integrand into the derivative of the given expression.
The integrand after this is:
$$\dfrac{cosec^2x cos^{2020}x - 2020cos^{2020}x}{cos^{4040}x}$$
=$$\dfrac{d/dx(-cotx) cos^{2020}x - 2020cos^{2019}x*cosx}{cos^{4040}x}$$
=$$\dfrac{d/dx(-cotx) cos^{2020}x - 2020cos^{2019}x*(-sinx)(-cotx)}{cos^{4040}x}$$
=$$\dfrac{d/dx(-cotx) cos^{2020}x - d/dx(cos^{2020}x)(-cotx)}{cos^{4040}x}$$
=$$\dfrac{d}{dx} \dfrac{-cotx}{cos^{2020}x}$$
(from the quotient rule $\dfrac{vu'-uv'}{v^2}= (u/v)'$)
From the fundamental theorem of calculus, the integral is therefore:
$$\dfrac{-cotx}{cos^{2020}x}$$
Which means $A= -1$, $B=1$, $f= cot x$ . The value of the expression is therefore $2+(-1)=1.$
