0
$\begingroup$

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?

EDIT

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.

$\endgroup$
  • 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Hint:

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...

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

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.