Question

( int frac{cos (2 x)-1}{cos (2 x)+1} d )
( =int-frac{sin ^{2} x}{cos ^{2} x} d x )
( operatorname{arct}-int frac{sin ^{2} x}{cos ^{2} x} d x )
( =-int tan ^{2} x d x )
( =-intleft(operatorname{suc}^{2} x-1right) d x )
( =-int sec ^{2} x d x+int 1 d x )
( =-tan x+x+c )
( left[because int sec ^{2} x d x=tan xright] )
( =x-tan x+c )
I then
( (c) )

# 7 dx = rcos 2x -1 Jcos 2x +1 (A) tan x - x+c (B) x + tan x + c X – tan x + c (D) - X - cot x + c

Solution