Question

evaluate the definite integral

Answer 1

\[\int\limits_{0}^{\pi/16} x(\sec(4x))^2dx\]

Answer 2

u = x
du = dx

dv = sec(4x)^2

1/4v = tan(4x)?? chain rule?

Answer 3

\[\frac{xtan(4x)}{4} - \int\limits \tan(4x) dx\]

Answer 4

\[\frac{xtan(4x)}{4} - -\frac{1}{4}\ln|\cos4x| + c ???????????\]

Answer 5

\[\frac{xtan(4x)}{4} +\frac{1}{4}\ln|\cos4x| + c ???????????\]

Answer 6

\[\frac{xtan(4x)}{4}\ln|\cos4x| + c ???????????\]

Answer 7

since \(\large dv=sec^2(4x)dx \), then \(\large v=\int sec^2(4x)dx=\frac{1}{4}tan(4x) \)

Answer 8

it's not 4tan(4x)? I thought it's the chainr ule...

Answer 9

\[\large [\frac{1}{4}tan(4x)]'=\frac{1}{4}[tan(4x)]'=\frac{1}{4}\cdot sec^2(4x)\cdot [4x]'=sec^2(4x) \]

Answer 10

chain rule for derivatives... i've never heard of the chain rule for integrals...

Answer 11

Me either.... I was just basing it off of what Turing and I did in my last problem....

Answer 12

um... :'(

Answer 13

your IBP procedure looks ok... it's just the part where you calculated v is incorrect...

Answer 14

I was just saying that a u-sub is like the opposite of the chain rule
there is no chain rule for integrals, you said that

Answer 15

@KonradZuse

Answer 16

oh okay so we didn't do a chain rule last time then huh... We just did u-sub..?

Answer 17

oic now.... !!!!!!!!!!!!!!!!!

Answer 18

Getting all my shizz mixed up :)

Answer 19

actually even the way I did it shouldn't it still be correct haha?

Answer 20

you dropped a factor of 1/4 from the second integral

Answer 21

besides that it's fine

Answer 22

oh wasn't looking at that haha :P.

Answer 23

\[\frac{xtan(x)}{4} + \frac{1}{8}\ln|\cos(4x)| + c\]

Answer 24

I mean nope

Answer 25

you're a trickster....

Answer 26

I obviously had to look twice, so you should catch your mistake as I did
you are so close

Answer 27

hmmmmmmmmmmmmmmmmmm

Answer 28

I'm gonna say that's totally your job to find your very silly algebra error. Good luck!

Answer 29

I hope it's not that + 1/8...?

Answer 30

that is where the problem is

Answer 31

well it's normally - int and when you convert tan it's -ln|cos(u)| +c

Answer 32

all that part is fine

Answer 33

unless your'e talking about the 1/8.. Which I thought with u sub it was 1/4 and the other 1/4 wouldbe 1/8.

Answer 34

oh really?

Answer 35

oh wait 1/16

Answer 36

:P

Answer 37

:P

Answer 38

I always do the dumbest algebra mistakes LOL!

Answer 39

now the fun part....

Answer 40

we all do

Answer 41

\[[\frac{\frac{\pi}{16}\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\cos(4\frac{\pi}{16})|] - [\frac{0\tan(0)}{4} + \frac{1}{16}\ln|\cos(0)|]\]

Answer 42

*yawn...

Answer 43

Ok I'm done here LOL!

Answer 44

wanna go simplify your other question to prove that it gives the same answer?

Answer 45

or are you done there too?

Answer 46

I feel like that one is basically done, just have to substitute and shizz... For this one I'm going to finish it up..

Answer 47

0tan(0)/4 should be 0?

Answer 48

yeah

Answer 49

2nd part is laso 0

Answer 50

cos0=1 and ln1=0 so all that on the right is 0

Answer 51

ugleh.

Answer 52

should I leave the first part like we did last night?

Answer 53

cos(pi/4)=sqrt(2)/2

Answer 54

ln|sqrt(2)/2|=who gives a crap?

Answer 55

;)

Answer 56

so yeah, I would leave that for my calculator lol

Answer 57

I should simpify to sqrt(2)/2 right?

Answer 58

\[[\frac{\frac{\pi}{16}\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\frac{\sqrt{2}}{2}|] - 0\]

Answer 59

I dunno, I mean... who cares?
do you want to use double-angle formulas to simplify tan(pi/16) too?
that seeems crazy

Answer 60

hahaha nty....

Answer 61

you could do some clever things with it, but which is the "simplest" is subjective

Answer 62

\[\frac{\frac{\pi}{16}\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\cos(\frac{\pi}{4})| - 0\]

Answer 63

haha that didn't work BOO!

Answer 64

ok so the practice version wants a decimal... REAQLLY! WTF ASK FOR IT...

Answer 65

\[\frac{\frac{\pi}{16}\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\frac{\sqrt{2}}{2}|=\frac\pi{16}\left(\frac14\tan(\frac\pi{16})-\frac12\ln2\right)\]\[=\frac\pi{64}\left(\tan(\frac\pi{16})-2\ln2\right)\]a little nicer

Answer 66

oh that pi is in the wrong place though

Answer 67

\[\frac{\frac{\pi}{16}\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\frac{\sqrt{2}}{2}|=\frac1{16}\left(\frac\pi4\tan(\frac\pi{16})-\frac12\ln2\right)\]\[=\frac1{64}\left(\pi\tan(\frac\pi{16})-2\ln2\right)\]

Answer 68

wolfram doesn't even like it... LOL

Answer 69

writing theanswer in the prettiest possible way is not wolfram's specialty

Answer 70

it just chugs out a number, or some systematic method of simplification

Answer 71

Maybe I'll enter your simpificatin and see if it likes it :P

Answer 72

omfg same answer....

http://www.wolframalpha.com/input/?i=1%2F64+%28pi+*+tan%28pi%2F16%29+-+2ln%7C2%7C%29

http://www.wolframalpha.com/input/?i=1%2F64+%28pi+*+tan%28pi%2F16%29+-+2ln%7C2%7C%29

Answer 73

yay

Answer 74

I'm scared to enter that lOL. Idk how many digits like 4...?

Answer 75

ok so -0.0119 didn't work...

Answer 76

=π/16(1) + 1/4ln sqrt{2}/2 = π/16 − 1/8 ln(2) ≈ 0.1097

From the practice version...

Answer 77

why is theirs positive.....

Answer 78

should it be 0.0119?

Answer 79

ok it whould be tan 4x, right, so you should have tan(pi/4)=1

Answer 80

so there's one mistake

Answer 81

question was \[\int\limits_{0}^{\pi/8} x(\sec(2x))^2dx\]

Answer 82

but you should still then have tan(2x) which is still tan(pi/4)

Answer 83

oh creap I messed that up :P

Answer 84

http://www.wolframalpha.com/input/?i=1%2F64+%28pi+*+tan%28pi%2F16%29+-+2ln%7C2%7C%29

Answer 85

\[\frac{\frac{\pi}4\tan(\frac{\pi}{16})}{4} + \frac{1}{16}\ln|\frac{\sqrt{2}}{2}|=\frac1{16}\left(\frac\pi4\tan(\frac\pi4)-\frac12\ln2\right)\]\[={\left(\pi-2\ln2\right)\over64}\]now it looks okay

Answer 86

WOOOOOOOOOOOOT! and we are done with that FOREVER.....

Answer 87

but you must walk away with a deeper knowledge... which I'm sure happened :)

Answer 88

Yeah I've learned a lot today :D.

Answer 89

ON TO THE NEXT QUESTION!

Answer 90

To learn more of course :P

Answer 91

oh joy :P