# "We would like to know [I assume you meant sin, instead of sen]:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad int \ { x^2 dx } / sin^2 ( 2 x^3 + 5 ) \ . #
# "One way to start, is to make a substitution" #
# "Let:" \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad u \ = \ 2 x^3 + 5. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (1) #
# "Compute:" \qquad \qquad \qquad \qquad \quad \quad du \ = \ 6 x^2 dx. #
# "Solve for" \ \ dx \ \":" \qquad \qquad \qquad \quad dx \ = \ { du } / { 6 x^2 }. #
# "Put" \ \ dx \ \ "back into original integral:" #
# \qquad \qquad \qquad \quad int \ { x^2 dx } / sin^2( 2 x^3 + 5 ) \ = \ int \ { x^2 } / sin^2( 2 x^3 + 5 ) \cdot { du } / { 6 x^2 }. #
# "Simplify in" \ \ x \ \ "as far as possible:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ int \ { color(red)cancel{x^2} } / sin^2( 2 x^3 + 5 ) \cdot { du } / { 6 color(red)cancel{x^2} } #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 int \ { du } / sin^2( 2 x^3 + 5 ). #
# "Express remaining integral completely in terms of" \ \ u. \ "This" #
# "can sometimes be difficult. But here, after recalling the" #
# "substitution eqn. (1), it is immediate [by design !!]:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 int \ { du } / sin^2( u ). #
# "Integrate new integral:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 int \ ( 1 / sin( u ) )^2 du #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 int \ [ csc( u ) ]^2 du. #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 int \ csc^2( u ) du. #
# "Recalling the basic derivative:" \ \ [ cot( \theta ) ]' = - csc^2( \theta ), "we finish the integration:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/6 \ [ - cot( u ) ] + C #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ - 1/6 \ cot( u ) + C. #
# "Now convert this result back into terms of" \ \ x. \ "This " #
# "is immediate, again using the substitution eqn. (1):" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ - 1/6 \ cot( 2 x^3 + 5 ) + C. #
# "And this finishes the work. So we have the answer:" #
# \qquad \qquad \qquad \quad int \ { x^2 dx } / sin^2( 2 x^3 + 5 ) = - 1/6 \ cot( 2 x^3 + 5 ) + C. #
