Question

Integral calc. is not my friend. Wondering if there is an easier way to solve the following integral than knowing/deriving the antiderivatives of csc(x) and sec(x)...

Answer 1

\[\int\limits_{}^{}\frac{ sinx+cosx }{ \sin(2x)}\]

Answer 2

\[\int\limits_{}{}\frac{ sinx+cosx }{ 2sinxcosx }\]
\[\int\limits_{}^{} \frac{ sinx }{ 2sinxcosx }+\int\limits_{}^{}\frac{ cosx }{ 2sinxcosx }\]
\[1/2\int\limits_{}^{}(cscx+secx)\]

Answer 3

dx everywhere

Answer 4

write sin2x as 1- (sinx - cos x)^2 and substitute sin x- cos x as t and sin x +cos x will become dt