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.