Question

Calculus II- Determine the integral by substitution
[see attachment]

Answer 1

\[\large \int\limits \cot\left(\color{royalblue}{2x}\right)\left(\color{orangered}{2dx}\right)\]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 \(\large du\)?

Answer 2

would du then just be 2

Answer 3

\(\large \color{orangered}{du=2dx}\)

yes good, don't forget the differential dx though.

Answer 4

\[\large \frac{du}{dx}=2 \qquad \rightarrow \qquad du=2dx\]

Answer 5

\[\large \int\limits\limits \cot\left(\color{royalblue}{2x}\right)\left(\color{orangered}{2dx}\right)\qquad\rightarrow\qquad \large \int\limits\limits \cot\left(\color{royalblue}{u}\right)\left(\color{orangered}{du}\right)\]

Answer 6

Do you understand how to solve it from here?
Converting to sines and cosines will help.

Answer 7

Hmm maybe we should have done that before applying our substitution..

Answer 8

thats what im confused about

Answer 9

\[\large \int\limits 2 \cot(2x)dx \qquad = \qquad \int\limits \frac{2\cos(2x)}{\sin(2x)}dx\]Understand that part?
I used a cotangent identity.

Answer 10

yes since tan is sinx/cosx

Answer 11

\[\large \int\limits\frac{\color{orangered}{2\cos(2x)dx}}{\color{royalblue}{\sin(2x)}}\]So this time, our substitution will be a little different.
We want the blue term to be our \(\large u\).

Answer 12

so if u was sin(2x), would it du be -cos value?

Answer 13

sine when differentiated produces cosine, not negative cosine, remember? :)

Answer 14

>.< nevermind

Answer 15

\[\large \color{royalblue}{u=\sin(2x)}\]\[\large \frac{du}{dx}=\cos(2x)(2x)'\]Due to the chain rule we have to multiply by the derivative of the inner function 2x.

Answer 16

\[\large \frac{du}{dx}=\cos(2x)(2)\]

Answer 17

"Moving" the dx to the other side gives us,\[\large \color{orangered}{du=2\cos(2x)dx}\]

Answer 18

Confused by any of that?

Answer 19

not yet...im still following what your saying...

Answer 20

i answered a problem for 8tan(8x) but cot is messing me up

Answer 21

would it go in 1/u = ln abs value +C? the same way?

Answer 22

\[\large \int\limits\frac{\color{orangered}{2\cos(2x)dx}}{\color{royalblue}{\sin(2x)}} \qquad=\qquad \int\limits\frac{\color{orangered}{du}}{\color{royalblue}{u}}\]

\(\large =\qquad \ln |u|+C\)

Yes very good.
Your last step would simply be to undo your u-substitution.

Answer 23

ln |sin(2x)| + C

Answer 24

Yay, good work

Answer 25

thank you very much