I cannot make an arctan because there is a sine in the numerator

[James Brady]

$\int\frac{2x^2sin(4x)}{1 + x^6}dx$

The solution should be substitution method... So far I've set $u = x^3$, and made some progress trying to make the integral ready to become an arctan:


$\frac{2}{3}\int\frac{sin(4x)}{1 + u^2}du$

The set up would be fine if it were not for the sine term in the numerator. Substituting the entire denominator doesn't really help the situation and I haven't been able to split the numerator either. Wolfram Alpha says it takes too long to compute so I haven't been able to see the solution either.

 

[lurflurf]

That is a fairly messy one, integrals.com gives

-I
2 ((-- (I CosIntegral[-4 I + 4 x] Sinh[4] -
2

Cosh[4] SinIntegral[4 I - 4 x])) /

1/6 1/6 5/6
((-I + (-1) ) (I + (-1) ) (I - (-1) )

5/6
(I + (-1) )) +

1/6 1/6
((-1) (CosIntegral[-4 (-1) + 4 x]

1/6
Sin[4 (-1) ] -

1/6 1/6
Cos[4 (-1) ] SinIntegral[4 (-1) - 4 x]))

1/6 1/6
/ (2 (-I + (-1) ) (I + (-1) )

1/6 5/6 1/6 5/6
((-1) - (-1) ) ((-1) + (-1) )) +

5/6 5/6
((-1) (CosIntegral[-4 (-1) + 4 x]

5/6
Sin[4 (-1) ] -

5/6 5/6
Cos[4 (-1) ] SinIntegral[4 (-1) - 4 x]))

5/6 1/6 5/6
/ (2 (I - (-1) ) ((-1) - (-1) )

5/6 1/6 5/6
(I + (-1) ) ((-1) + (-1) )) +

I
(- (-I CosIntegral[4 I + 4 x] Sinh[4] +
2

Cosh[4] SinIntegral[4 I + 4 x])) /

1/6 1/6 5/6
((-I - (-1) ) (-I + (-1) ) (-I + (-1) )

5/6
(I + (-1) )) +

1/6 1/6
((-1) (-(CosIntegral[4 (-1) + 4 x]

1/6
Sin[4 (-1) ]) +

1/6 1/6
Cos[4 (-1) ] SinIntegral[4 (-1) + 4 x]))

1/6 1/6
/ (2 (-I - (-1) ) (I - (-1) )

1/6 5/6 1/6 5/6
(-(-1) + (-1) ) ((-1) + (-1) )) -

5/6 5/6
((-1) (-(CosIntegral[4 (-1) + 4 x]

5/6
Sin[4 (-1) ]) +

5/6 5/6
Cos[4 (-1) ] SinIntegral[4 (-1) + 4 x]))

5/6 5/6
/ (2 (-I - (-1) ) (I - (-1) )

1/6 5/6 1/6 5/6
(-(-1) - (-1) ) ((-1) - (-1) )))

Edited to add: Do you want the 0 to infinity integral or indefinite?

 

[HallsofIvy]

$\int\frac{2x^2sin(4x)}{1 + x^6}dx$

The solution should be substitution method... So far I've set $u = x^3$, and made some progress trying to make the integral ready to become an arctan:


$\frac{2}{3}\int\frac{sin(4x)}{1 + u^2}du$

This itself is incorrect. You cannot have an "x" in the integral with respect to "u".

The set up would be fine if it were not for the sine term in the numerator. Substituting the entire denominator doesn't really help the situation and I haven't been able to split the numerator either. Wolfram Alpha says it takes too long to compute so I haven't been able to see the solution either.

 

[lurflurf]

^I posted the result from integrals.com above
Complex partial fractions gives terms like
sin(x)/(x+a)
real partial fractions gives terms like
sin(x)/((x-a)^2+b^2)
In either case much algebra and integration by parts gives a result in terms of sin cos sinh cosh type integrals

if

$$\mathop{I}(a) = \int \! \frac{\sin (4 \, x)}{x^2+a \sqrt{3}+1} \, \mathop{dx}$$

The given integral can be written as

(1/3) I(-sqrt(3))+(-2/3) I(0)+(1/3) I(sqrt(3))

http://www.wolframalpha.com/input/?i=integral+sin(4x)/(x^2+a+sqrt(3)+x+1)
http://integrals.wolfram.com/index.jsp?expr=2x^2+Sin[4x]/(1+x^6)&random=false

 

[James Brady]

This itself is incorrect. You cannot have an "x" in the integral with respect to "u".

Yes, I know... I was thinking I could rewrite the 4x in terms of u, but seeing from the answer above, this is way beyond my skill level either way. The actual problem was taking the definite integral from -pi/2 to pi/2. So the professor was just showing us that even with really complex integrals like this, with symmetry, we can still get the answer, which in this case is 0. Haha, that's one of those "oh yeah..." moments.

Thanks all.

 

[haruspex]

The actual problem was taking the definite integral from -pi/2 to pi/2.

Please don't omit information like that when you post new questions. It wastes a lot of time and effort.

 

[James Brady]

Please don't omit information like that when you post new questions. It wastes a lot of time and effort.

Do you even lift bro?

 

[Millennial]

Do you even lift bro?

Haruspex is more than correct here. With the information of limits, it is easy to solve the question.
With definite integrals of messy functions, you will first want to examine the function itself. Is there any particular property of the function that would make integration easier?

Hint: Do you know what even and odd functions are?

 

[James Brady]

Right? After several hours of integration. I can definitely say that taking a few moments to analyze for symmetry is worth the time.