0
$\begingroup$

I have nearly solved this question but I'm getting a different answer when substituting limits:

$∫_0^2 \left[\sin⁡(2x)+\sec^2 (\frac{x}{2})\right]\,dx $

integrated to: $-\frac{1}{2}\cos(2x)+2\tan(\frac{x}{2})+c$

substituting limits:

($-\frac{1}{2}\cos(4)+2\tan(1))-(-\frac{1}{2}\cos(0)+2\tan(0))+c$

simplify: $-\frac{1}{2}\cos(4)+2\tan(1)+\frac{1}{2}$

For me this evaluates to $0.036128...$, but the solution is actually $3.94163$...

Is there something obvious I'm missing?

$\endgroup$
4
  • 1
    $\begingroup$ When you're doing a definite integral, the constant of integration cancels out. Also, is your calculator in the right mode? Degrees versus radians? $\endgroup$ May 22, 2018 at 17:32
  • $\begingroup$ Did you use degrees or radians?? $\endgroup$ May 22, 2018 at 17:32
  • $\begingroup$ Thanks. So is 0.036128 degrees equally correct? Also, does that mean the limits should be stated in radians? How would one know whether an integral should be reported in radians or degrees? $\endgroup$
    – Paddington
    May 23, 2018 at 11:52
  • $\begingroup$ Thanks again. I just found this guide which explains it nicely. $\endgroup$
    – Paddington
    May 23, 2018 at 12:01

1 Answer 1

5
$\begingroup$

Did you use Degrees or radians while calculating it, because it is correct for radians.

radians

enter image description here

degrees

enter image description here

$\endgroup$

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.