anonymous
  • anonymous
Integration by substitution
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amoodarya
  • amoodarya
where is question?
anonymous
  • anonymous
general integral with the function sin(2x) dx My work: u = 2x du/dx = 2 du = 2dx 1/2 integral (2sin(2x)dx) 1/2 integral (sinu * du) 1/2sin(2x) + C but that's not right i believe.. lol
anonymous
  • anonymous
is it supposed to be cos instead of sin?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
keep your derivatives in mind, what happens if you derive -cos(x)?
anonymous
  • anonymous
sin
anonymous
  • anonymous
exactly.
anonymous
  • anonymous
oh i just checked the answer sheet -_- it says cos...
anonymous
  • anonymous
1/2 (cos(2x)) + C
anonymous
  • anonymous
hmm, well that's not the answer in my opinion, because if you derive that you get: \[\Large - \frac{1}{2} \sin(2x) \cdot 2 = - \sin(2x) \]
anonymous
  • anonymous
oh... i didn't get that at all. idk, the given answer is 1/2cos(2x) + C
anonymous
  • anonymous
is my u sub wrong?
anonymous
  • anonymous
No your substitution is perfect, it's more the way you integrated, you kept the sin function, which shouldn't be the case when you integrate, because when you differentiate you want to obtain the integrand again \[F(x)\prime =f(x) \]
anonymous
  • anonymous
OHHHHHH snap I get it! lol
tkhunny
  • tkhunny
Seriously? Why would you EVER use substitution for a mere constant? It's just craziness. Speculate \(\int \sin(2x)\;dx = -\cos(2x)\;+\;C\) Check \(\dfrac{d}{dx}(-\cos(2x)) = 2\cdot \sin(2x)\) -- Oops, we missed a constant. Solve \(\int \sin(2x)\;dx = -\dfrac{1}{2}\cos(2x)\;+\;C\) -- Done. On the other hand: \(\int \sin(2x)\;dx = \int 2\sin(x)\cos(x)\;dx\) -- Now, THERE'S a candidate for Substitution.
anonymous
  • anonymous
because I've been doing exponential functions all day -_- so i kept that in mind that i shouldn't change it. But just to make sure... i should integrate the function after i derive the u correct?
anonymous
  • anonymous
oh the constant part i have no prob with hahaha general integrals = add or subtract c lol
tkhunny
  • tkhunny
I wish I could tell what "derive" means. I have not found it acceptable to use it as the verb form for finding a derivative.
anonymous
  • anonymous
i'm not sure if that is sarcasm because i was tempted to give you the definition lolll but thanks!
tkhunny
  • tkhunny
No, not sarcasm, just discouragement. I don't recommend that usage. It just isn't generally in use.
anonymous
  • anonymous
oh here it is. as long as we know the answer and the concept, it's all good.
anonymous
  • anonymous
exactly.
tkhunny
  • tkhunny
And perhaps that we managed to learn something else along the way. Good work.

Looking for something else?

Not the answer you are looking for? Search for more explanations.