anonymous
  • anonymous
Calculus II- Determine the integral by substitution [see attachment]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
zepdrix
  • zepdrix
\[\large \int\limits \cot\left(\color{royalblue}{2x}\right)\left(\color{orangered}{2dx}\right)\]There's a reason I wrote it this way, maybe it will make sense in a moment. \[\large \color{royalblue}{u=2x}\] What do you get for your \(\large du\)?
anonymous
  • anonymous
would du then just be 2

Looking for something else?

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

More answers

zepdrix
  • zepdrix
\(\large \color{orangered}{du=2dx}\) yes good, don't forget the differential dx though.
zepdrix
  • zepdrix
\[\large \frac{du}{dx}=2 \qquad \rightarrow \qquad du=2dx\]
zepdrix
  • zepdrix
\[\large \int\limits\limits \cot\left(\color{royalblue}{2x}\right)\left(\color{orangered}{2dx}\right)\qquad\rightarrow\qquad \large \int\limits\limits \cot\left(\color{royalblue}{u}\right)\left(\color{orangered}{du}\right)\]
zepdrix
  • zepdrix
Do you understand how to solve it from here? Converting to sines and cosines will help.
zepdrix
  • zepdrix
Hmm maybe we should have done that before applying our substitution..
anonymous
  • anonymous
thats what im confused about
zepdrix
  • zepdrix
\[\large \int\limits 2 \cot(2x)dx \qquad = \qquad \int\limits \frac{2\cos(2x)}{\sin(2x)}dx\]Understand that part? I used a cotangent identity.
anonymous
  • anonymous
yes since tan is sinx/cosx
zepdrix
  • zepdrix
\[\large \int\limits\frac{\color{orangered}{2\cos(2x)dx}}{\color{royalblue}{\sin(2x)}}\]So this time, our substitution will be a little different. We want the blue term to be our \(\large u\).
anonymous
  • anonymous
so if u was sin(2x), would it du be -cos value?
zepdrix
  • zepdrix
sine when differentiated produces cosine, not negative cosine, remember? :)
anonymous
  • anonymous
>.< nevermind
zepdrix
  • zepdrix
\[\large \color{royalblue}{u=\sin(2x)}\]\[\large \frac{du}{dx}=\cos(2x)(2x)'\]Due to the chain rule we have to multiply by the derivative of the inner function 2x.
zepdrix
  • zepdrix
\[\large \frac{du}{dx}=\cos(2x)(2)\]
zepdrix
  • zepdrix
"Moving" the dx to the other side gives us,\[\large \color{orangered}{du=2\cos(2x)dx}\]
zepdrix
  • zepdrix
Confused by any of that?
anonymous
  • anonymous
not yet...im still following what your saying...
anonymous
  • anonymous
i answered a problem for 8tan(8x) but cot is messing me up
anonymous
  • anonymous
would it go in 1/u = ln abs value +C? the same way?
zepdrix
  • zepdrix
\[\large \int\limits\frac{\color{orangered}{2\cos(2x)dx}}{\color{royalblue}{\sin(2x)}} \qquad=\qquad \int\limits\frac{\color{orangered}{du}}{\color{royalblue}{u}}\] \(\large =\qquad \ln |u|+C\) Yes very good. Your last step would simply be to undo your u-substitution.
anonymous
  • anonymous
ln |sin(2x)| + C
zepdrix
  • zepdrix
Yay, good work
anonymous
  • anonymous
thank you very much

Looking for something else?

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