If f(x)=cosx and g(x)=sinx then ∫logf(x)(f(x))2dx is
f(x)(logf(x)+1)+x+C
g(x)f(x)(logf(x)+1)+x22+C
f(x)g(x)(logg(x)+1)+x+C
g(x)f(x)[logf(x)+1]−x+C
I=∫logcosxcos2xdx=∫sec2xlogcosxdx=(tanx)logcosx−∫tanx−sinxcosxdx=(tanx)logcosx+∫sec2−1dx=(tanx)[logcosx+1]−x+C
=g(x)f(x)[logf(x)+1]−x+C