So I have to solve this integral $$ \int x\cos(1+x^2)\sin(1-x^3)\; dx $$

I know that there are ways to solve trigonometric integrals depending on do the sine and cosine have odd or even degree or if they are represented in some of the standard forms, and if nothing, some of them can be solved using trigonometric identities to simplify the expression to make it easy to integrate. But, in this case I don't have a good idea, I tried to use the formula

$$\sin\alpha\cos\beta= \frac{1}{2}(\sin(\alpha + \beta)+\sin(\alpha-\beta))$$

but it turns out it isn't helpful because, then I have the following integral

$$\frac{1}{2} \int x(\sin(2+x^2-x^3)+\sin(x^2+x^3))\; dx $$

which doesn't seem any less complicated than the one I had at the beginning. Any ideas?

  • 3
    $\begingroup$ What makes you think there is an indefinite integral? $\endgroup$ – almagest Mar 18 '16 at 16:57
  • $\begingroup$ @almagest i am not sure what you mean, are you trying to say that function under integral has no primitive function in the set of elementary functions? $\endgroup$ – cdummie Mar 18 '16 at 17:01
  • 2
    $\begingroup$ Please share your anti-derivative $\endgroup$ – Mark Viola Mar 18 '16 at 18:23
  • 2
    $\begingroup$ @Dr.MV i second this resquest $\endgroup$ – tired Mar 18 '16 at 19:09
  • 1
    $\begingroup$ That isn't the correct antiderivative. There is no cubic term to be found. $\endgroup$ – Mark Viola Mar 19 '16 at 15:06

I will make an educated guess that this integral is supposed to be:

$$\int x\cos(1+x^2)\sin(1-x^2)\; dx$$

Then, by using the OP's suggestion, we tranform the product into a sum:

$$\frac{1}{2} \int x \sin 2 ~dx-\frac{1}{2} \int x \sin (2x^2)dx=\frac{\sin 2}{4}x^2+\frac{1}{8} \cos (2x^2)+C$$

The 'antiderivative' the OP provided in the comments leads to the integral:

$$\frac{-1}{4}x^2 \cos2x^2 + \frac{1}{8} \sin2x^2 + C=\int x^3 \sin (2x^2) dx$$

This is obviously not the same integral.

The first integral can be easily generalized to an arbitrary degree:

$$\int x^{n-1} \cos(1+x^n)\sin(1-x^n)\; dx=$$

$$=\frac{1}{2} \int x^{n-1} \sin 2 ~dx-\frac{1}{2} \int x^{n-1} \sin (2x^n)dx=\frac{\sin 2}{2n}x^n+\frac{1}{4n} \cos (2x^n)+C$$

  • $\begingroup$ What if it is $x^3$ instead of $x^2$? Is it proven that it cannot be solve? @Yuriy S $\endgroup$ – Tony Ma Apr 6 '18 at 9:00
  • $\begingroup$ @TonyMa, it's very likely that a closed form antiderivative in terms of elementary functions doesn't exist in this case. See en.wikipedia.org/wiki/Risch_algorithm for further reference $\endgroup$ – Yuriy S Apr 6 '18 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.