∫xtan2(3x2)dx
Note that if
d(3x2)=(3x2)′dx=6xdx
then
xdx=61⋅d(3x2)
∫x⋅tan2(3x2)dx=
∫tan2(3x2)⋅xdx=
∫tan2(3x2)⋅61⋅d(3x2)=
61⋅∫tan2(3x2)⋅d(3x2)
Further substitute
tan(3x2)=t
Hence
3x2=arctan(t)
d(3x2)=d(arctan(t))=(arctan(t))′dt=t2+11dt
Continue solving
61⋅∫tan2(3x2)⋅d(3x2)=
61⋅∫t2⋅t2+11dt=
61⋅∫t2+1t2dt=
61⋅∫t2+1t2+1−1dt=
61⋅∫(t2+1t2+1−t2+11)dt=
61⋅∫(1−t2+11)dt=
61⋅(t−arctan(t))+c= <return to variable x: t=tan(3x2) >
61⋅(tan(3x2)−arctan(tan(3x2)))+c=
61⋅(tan(3x2)−3x2)+c=
61tan(3x2)−61⋅3x2+c=
61tan(3x2)−2x2+c
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