heys, here's my attempt:
[ 2 / (1 + sin 4x) ] + [ 2 / ( 1 -
sin 4x) ] dx
= 2 [(1-sin 4x + 1 + sin 4x)/(1-sin^2 4x)] dx
(factorise out the 2 on the numerator first, then combine the two
fractions together, denominator simplified using (a+b)(a-b)=(a^2 -
b^2))
= 4 [1/(1-sin^2 4x)] dx
yea. i'm not sure what A math scope is so, if there's a formula in
your book that somehow corresponds to the expression then please
use it. sorry that's all i can help.
There is no need for you to apologise, as long as you have tried
your very best to help out.
[ 2 / (1 + sin 4x) ] + [ 2 / ( 1 -
sin 4x) ]
= [ 2(1 - sin 4x) / (1 + sin 4x)( 1 - sin 4x) ] + [
2(1 + sin 4x) / (1 + sin 4x)(1 - sin 4x) ]
= [ 2 - 2sin 4x + 2 + 2sin 4x ] / (1 + sin 4x)( 1 - sin 4x)
= 4 / [ 1 - (sin 4x)^2 ]
= 4 / [ (sin4x)^2 + (cos4x)^2 - (sin4x)^2 ]
= 4 / (cos4x)^2
= 4
(sec4x)^2
Integrate this, and you will get tan 4x, and you should be able
to continue from here.