"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.