Question #ec3c9

2 Answers
Feb 19, 2018

#I=-1/6cot(2x^3+5)+C#

Explanation:

We want to solve

#I=intx^2/sin^2(2x^3+5)dx#

Make a substitution #u=2x^3+5=>(du)/dx=6x^2#

#I=intx^2/sin^2(u)1/(6x^2)du=1/6intcsc^2(u)du#

Recall #(cot(x))'=-csc^2(x)#

#I=-1/6cot(u)+C#

Substitute back #u=2x^3+5#

#I=-1/6cot(2x^3+5)+C#

Feb 19, 2018

# \qquad \qquad \qquad \quad \ \ int \ { x^2 dx } / sin^2( 2 x^3 + 5 ) = - 1/6 \ cot( 2 x^3 + 5 ) + C. #

Explanation:

# "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. #