Question

Integrate cos2x log(cosx+sinx/cosx-sinx)

Answer 1

∫ cos(2x) * ln( sin(x) ) <===> by parts

u = ln( sin(x) ) . . . . . . . . dv = cos(2x)
du = (1/sin(x)) cos(x) .. . .v = (1/2) sin(2x)

u * v - ∫ v * du
ln( sin(x) ) * (1/2) sin(2x) - ∫ (1/2) sin(2x) * (cos(x)/sin(x))
ln( sin(x) ) * (1/2) sin(2x) - ∫ (1/2) 2in(x)cos(x) * (cos(x)/sin(x))
ln( sin(x) ) * (1/2) sin(2x) - ∫ cos^2(x)
ln( sin(x) ) * (1/2) sin(2x) - ∫ (1/2) * (1 + cos(2x))
ln( sin(x) ) * (1/2) sin(2x) - ∫ (1/2) + (1/2)cos(2x)
ln( sin(x) ) * (1/2) sin(2x) - (1/2)x - (1/4)sin(2x) =====> from 0 to π/2

[ ln( sin(π/2) ) sin(2 * π/2) - ln( sin(0) ) * sin(2 * 0) ] - (1/2) * ( π/2 - 0 ) - (1/4) * ( sin(2 * π/2) - sin(2 * 0) ) ]

[ ln(1) sin(π) - ln(0) * sin(0) ] - (1/2) * π/2 - (1/4) * ( sin(π) - sin(0) ) ]
[ 0 * 0 - -∞ * 0 ] - π/4 - (1/4) * ( 0 - 0 )
[ 0 + ∞ * 0 ] - π/4 - 0
[ ∞ * 0 ] - π/4 <===== but here we have the Not determinant case ∞ * 0


lim ln( sin(x) ) sin(2x)
x--->0

ln( sin(x) ) / csc(2x) <===== apply L'Hopital Rules
[ cos(x)/sin(x) ] / [ - 2csc(2x)cot(2x) ]
[ cos(x)/sin(x) ] / [ - 2(1/sin(2x))( cos(2x)/sin(2x) ) ]
[ cos(x)/sin(x) ] / [ - 2(1/sin(2x))( cos(2x)/2sin(x)cos(x) ) ]
[ cos(x) ] / [ - (cos(2x)/sin(2x)cos(x) ) ]
[ cos(x) ] * [ - sin(2x)cos(x) / cos(2x) ]
[ - sin(2x)cos^2(x) / cos(2x) ] ======> lim as x--> 0

[ - sin(2 * 0)cos^2(0) / cos(2*0) ]
[ - 0 * 1 / cos(0) ]
[ 0 / 1 ] = 0


[ ∞ * 0 ] - π/4 <===== will be as

0 - π/4