I want to solve this:

$$\int 3x \cos x^2 \, dx$$

I get this answer:

$$ \frac{\sin 2x}{2}+\frac{\cos 2x}{4}+C $$

but the answer should be:

$$ \frac{3 \sin x^2}{2}+C $$

Am I doing anything wrong or is it possible to rewrite the solution in another way so that they are the same?


Here is my calculation:

$$3\int x\cos^2x = 3\int x\frac{1}{2}(1+\cos2x)dx=\frac{3}{2}\int x+x\cos2x dx=\frac{3x^2}{4}\int x\cos2x dx=\frac{\sin2x}{2}-\int \frac{\sin2x}{2}dx=\frac{\sin2x}{2}x+\frac{\cos2x}{24}$$

But as said, the assumption $\cos^2x=\cos x^2$ is clearly wrong.

  • 1
    $\begingroup$ Your version is not equivalent to the correct answer. $\endgroup$ – André Nicolas Dec 18 '14 at 20:59
  • $\begingroup$ To get the answer you are heading for, it would seem that the integral should be written as $$\int3x\cos\left(x^2\right)\,\mathrm{d}x$$ $\endgroup$ – robjohn Dec 18 '14 at 21:00
  • 2
    $\begingroup$ Show us the steps you used to get to your answer? Because then we can help identify some of the errors in your way :) $\endgroup$ – Chinny84 Dec 18 '14 at 21:03
  • 1
    $\begingroup$ @robjohn I know many math teachers and professors who hate the common shorthand of omitting the brackets/parentheses from trig functions for exactly this reason. $\endgroup$ – KSmarts Dec 18 '14 at 21:03
  • $\begingroup$ Updated with steps! $\endgroup$ – theva Dec 18 '14 at 22:21

HINT: Let $t=x^2$, $dt=2x\,dx$ and the rest should be simple.

So: yes, there is a mistake in your solution. Your answer cannot be identical with the correct one, because the former function is periodic and the latter not.

  • $\begingroup$ This looks like a perfectly good hint. Why the downvote? $\endgroup$ – robjohn Dec 18 '14 at 21:05
  • $\begingroup$ I suspect that I misunderstood the $cosx^2$ i interpreted it as cos^2x... $\endgroup$ – theva Dec 18 '14 at 21:05
  • $\begingroup$ @theva even accounting for that you don't get the result you show..we can help if you show us then we can fix your issue. $\endgroup$ – Chinny84 Dec 18 '14 at 21:20


U-Substitution first

$$ u = x^2 $$ Therefore $$du = 2xdx $$ then $$ \frac{du}{2x} = dx$$

$$ \int 3x \cos x^2 \, dx =\int 3x \cos x^2 \frac{du}{2x} = \int \frac{3}{2} \cos u \, du = \frac{3}{2}\int \cos u\, du $$

Now you can figure integral of cos out by yourself...


$$ \int 3x \cos(x^2) \, dx = \frac{3}{2} \sin(x^2) + C $$

as can be checked by

$$ \left[\frac{3}{2} \sin(x^2) + C \right]' = \frac{3}{2} \, \cos(x^2)\, 2x = 3x \cos(x^2) $$

If you do not see it, you could use the substitution $u = x^2$: $$ \int 3x \cos(x^2) \, dx = \int 3x \cos(u) \, \frac{du}{2x} = \int \frac{3}{2} \cos(u) \, du= \frac{3}{2} \sin(u) + C = \frac{3}{2} \sin(x^2) + C $$


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