Question

integration by parts help

Answer 1

\[\int\limits_{}^{}8xsec ^{2}(2x)dx\]

Answer 2

i got

4xtan(2x)-4|sec(2x)+c
but its wrong, why

Answer 3

for integration by parts you would need to know
1) the integral of sec^2(2x)
2) the derivative of 8x.

Answer 4

u= 8x du=8 dx
dv= sec^2 (2x) v=(1/2)tan(2x)

(4xtan(2x))-[anti sign](4tan(2x) dx
pulled the four out
-4[anti-sign] tan(2x)
anti of tan is ln|sec x|
so
i got my answer that way

Answer 5

no need for u -sub.

Answer 6

Love to disconnect

Answer 7

Anyway

Answer 8

(Without any +C for now)
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\sec^2(x)dx=\tan(x)}\) right?
So when you have:
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\sec^2(2x)dx}\) you get, \(\large\color{slate}{(1/2)\tan(2x)}\)

Answer 9

And derivative of 8x is 8... (we know that I am sure :D )

Answer 10

Ok, now lets proceed.

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}8x\sec^2(2x)dx}\)

\(\large\color{slate}{8x(1/2)\tan(2x)-\displaystyle\int\limits_{~}^{~}8(1/2)\tan(2x)dx}\)

\(\large\color{slate}{4x\tan(2x)-4\displaystyle\int\limits_{~}^{~}\tan(2x)dx}\)

Answer 11

understanding so far?

Answer 12

yes that what i did right so far

Answer 13

yes, now you need to integrate the last integral

Answer 14

you can set u=2x, if you really feel uncomfortable working with the "2x" chain every time.
\(\large\color{slate}{I}\) though would just do it the way it is:

Answer 15

i got

8ln|sec(2x)|

Answer 16

what is the entire answer/

Answer 17

?

Answer 18

4xtan(2x)-8ln|sec(2x)|+c

Answer 19

let me check

Answer 20

\(\large\color{slate}{4x\tan(2x)-4\displaystyle\int\limits_{~}^{~}\tan (2x)dx}\)

\(\large\color{slate}{4x\tan(2x)-4\displaystyle\int\limits_{~}^{~}\sin (2x)~/~\cos(2x)dx}\)


\(\large\color{slate}{u=\cos(2x)}\)
\(\large\color{slate}{du=-2\sin(2x)~dx}\)


\(\large\color{slate}{4x\tan(2x)+2\displaystyle\int\limits_{~}^{~}1/u~~du}\)

\(\large\color{slate}{4x\tan(2x)+2\ln(u)}\)

Answer 21

and then sub u.

Answer 22

See your err?

Answer 23

hmmmmm puzzling, what was wrong doing u substitution with just tan, if you have time

Answer 24

yes, I have time, wat is exactly troubling you about this.

Answer 25

?

Answer 26

sorry lost connection

Answer 27

well i did
\[4\int\limits_{}^{}\tan(2x)dx \]
u=2x

du=2dx

du/2 = dx

\[4\int\limits_{}^{}\tan(u) \left(\begin{matrix}du \\ 2\end{matrix}\right)\]

multiplied the 2 across and got
\[8\int\limits_{}^{}\tan(u) du\]
tan(u) anti derivative is ln|sec(x)|

so i got

8ln|sex(2x)|

Answer 28

sec not sex lol

Answer 29

so the final should be

16 ln|sec(2x)|

Answer 30

Wouldn't curse my biggest enemy like this... cursed connection shall never be blessed

Answer 31

\(\large\color{blue}{4x\tan(2x)-4\displaystyle\int\limits_{~}^{~}\tan(2x)~dx=}\)


\(\large\color{blue}{4x\tan(2x)-4\displaystyle\int\limits_{~}^{~}\frac{\sin(2x)}{\cos(2x)} ~dx=}\)

\(\large\color{blue}{4x\tan(2x)+2\left[ -2\displaystyle\int\limits_{~}^{~}\frac{\sin(2x)}{\cos(2x)} ~dx \right] =}\)


\(\large\color{red}{u=\cos(2x)}\)
\(\large\color{green}{du=-2\sin(2x)~dx}\)

\(\large\color{blue}{4x\tan(2x)+2\left[ \color{green}{-2}\displaystyle\int\limits_{~}^{~}\frac{\color{green}{\sin(2x)}}{\color{red}{\cos(2x)}}~\color{green}{dx} \right] =}\)

Answer 32

Is this making sense?

Answer 33

yes

Answer 34

\(\large\color{blue}{4x\tan(2x)+2\left[ \color{green}{-2}\displaystyle\int\limits_{~}^{~}\frac{\color{green}{\sin(2x)}}{\color{red}{\cos(2x)}}~\color{green}{dx} \right] =}\)


\(\large\color{red}{u=\cos(2x)}\)
\(\large\color{green}{du=-2\sin(2x)~dx}\)


\(\large\color{blue}{4x\tan(2x)+2\left[ \displaystyle\int\limits_{~}^{~}\frac{\color{green}{du} }{\color{red}{u}}~\right] =}\)

Answer 35

\(\large\color{blue}{4x\tan(2x)+2\left[ \displaystyle\int\limits_{~}^{~}\frac{1 }{\color{red}{u}}\color{green}{du}~\right] =}\)

\(\large\color{blue}{4x\tan(2x)+2\displaystyle\int\limits_{~}^{~}\frac{1 }{\color{red}{u}}\color{green}{du} =}\)

\(\large\color{blue}{4x\tan(2x)+2\ln(\color{red}{u})+C}\)

\(\large\color{blue}{4x\tan(2x)+2\ln(\color{red}{~\cos(2x)~})+C}\)

Answer 36

See what is happening?

Answer 37

yes its a lot clear now, dont know how i missed that

Answer 38

there is no negative in front of the ln to make it a secant inside

Answer 39

that u=cos(x) is the definition of how to solve an integral of tan(x)

Answer 40

just like you would set u=sin(x) to solve integral of cot(x)`

Answer 41

yes, thank you so much. i know it took a long time and i apologize but really thank you thank you thank you

Answer 42

No need for apologies, yw.

Answer 43

I would integrate (by parts)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~} \frac{e^x}{x}~dx }\)

Answer 44

do you see the latex?

Answer 45

Anyways, just remember the thing about an integral of tan(x), (that u=cos(x) )

Answer 46

But keep track of +'s and -'s.

Answer 47

yw, bye