Question

integral from 0 to pi/15 of xtan^2(5x) dx

please help.

Answer 1

u=x and du=dx v=1/5tan(5x)-x I am setting up the problem now. uv-integral vdu

Answer 2

\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-\frac{ 1 }{ 5 }(\frac{ 1 }{ 5 }\ln |\sec(5x)|-x)dx\]

Answer 3

\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-x)dx\]

Answer 4

tan^2 = 1-sec^2 which might make life a little easier

Answer 5

i used that for substitution before i started integrating

Answer 6

i set dv = integral sec^2(5x)-1 dx

Answer 7

\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-x)dx| x=\pi/15 and x=0\]

Answer 8

x(1-s^2)

x - x s^2

x ints up to x^2/2

x s^2 tables out as

s^2
x t
-1 - ln cos

x^2/2 - xtan(x) - ln(cos(x))

looks right to me

Answer 9

of course the 5 has to fit somplace :)

Answer 10

\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-\frac{ x^2 }{ 2 })dx| x=\pi/15 and x=0\]

Answer 11

ok does this look right to start evaluating at now?

Answer 12

the dx at the end is bad notation

x^2/2 - xtan(5x)/5 - ln(cos(5x))/25 get me a negative on the mix .... so ive altered a sign someplace

Answer 13

oh yeah.. bad dx.. so used to writing it. ok .. here is my problem Amistre. when i distribute the pi/15... i get the wrong answer. the right half of the equation is correct.. -1/25 ln .... thats right.. but... the left side is wrong

Answer 14

after simplifying I will end up getting -pi(pi5sqrt(3)-pi/something)

Answer 15

\[\left[ \frac{ \pi }{75 }\sqrt{3}-\frac{ \pi^2 }{ 225 } \right]\]

Answer 16

let me check my derivative

x^2/2 - xtan(5x)/5 - ln(cos(5x))/25

x - tan(5x)/5 - xsec^2(5x) + tan(5x)/5

x - xsec^2(5x)

x(1-sec^2(5x)) = xtan^2(5x)

is there an error? the wolf says its negative

Answer 17

i see, tan^2 = sec^2 - 1

Answer 18

xtan^2(5x)

x(sec^2(5x)-1)

xsec^2(5x) - x

pi.tan(pi/3)/5.15 + ln(cos(pi/3))/25 - pi^2/2.15^2

0.tan(5.0)/5 + ln(cos(5.0))/25 - 0^2/2


sqrt(3).pi/5.15 + ln(1/2)/25 - pi^2/2.15^2

Answer 19

the answer i need to get to is
\[\frac{ \pi \sqrt{3} }{ 75 }-\frac{ 1 }{ 25 }\ln2-\frac{ \pi^2 }{ 450 }\]

Answer 20

5*15 = 75

2*15^2 = 450

ln(1/2) = -ln(2)

Answer 21

it is the pi*sqrt(3)/75 that i cannot figure out

Answer 22

pi * tan(pi/3) / (5*15)

since tan(pi/3) = sqrt(3) ....

Answer 23

\[\left[ \frac{ \pi }{75 }\sqrt{3}-\frac{ \pi^2 }{ 225 } \right]\]

what about the last term? how does that just disappear.

Answer 24

\[ \frac x5~tan(5x)\]
\[ \frac{\pi}{5*15}~tan(60^o)\]
\[ \frac{\pi\sqrt3}{75}\]

Answer 25

\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-\frac{ x^2 }{ 2 }\]

Answer 26

coming from this.. do you not distribute to the last term?

Answer 27

or should the -x not be in there? for uv it is x * 1/5 tan (5x) -x

Answer 28

\[=\pi/15(\frac{ 1 }{ 5 }\tan(5\pi/15)-\pi/15)\]

maybe this isnt right?

Answer 29

i think youve got an error in there someplace, i get an integral of
\[\frac x5tan(5x) +\frac1{25} ln[cos(5x)] - \frac{1}{2}x^2\]

Answer 30

at x=0 that all goes to zero, so its a focus on x=pi/15

Answer 31

isnt the formula uv-integral udv ?

Answer 32

thus if u = x and v = 1/5tan(5x)-x ... man i feel dumb lol

Answer 33

it is, which can be tabled as well
v
u sec^2(5x)
x tan(5x)/5 <-- multply
-1 -ln(cos(5x))/25 <-- multply
---------
sum

Answer 34

the stand alone x can be integrated on its one

Answer 35

\[\frac x5\tan(5x) +\frac1{25} \ln[\cos(5x)] - \frac{1}{2}x^2\]

Answer 36

in the first part of this... where is the other x?

Answer 37

that x^2/2 on the end is the result of integrates -x

Answer 38

\[\frac x5\tan(5x) \]

Answer 39

isnt that part of the vdu? isnt vdu 1/5tan(5x)-x?

Answer 40

\[x~tan^2(5x)=x(sec^2(5x)-1)=x~sec^2(5x)-x\]
sec^2 is easier to play with than tan^2
\[\int x~sec^2(5x)-x~dx\]
\[\int x~sec^2(5x)~dx-\frac12x^2\]
\[\frac15x~tan(5x)-\int \frac15tan(5x)~dx-\frac12x^2\]
\[\frac15x~tan^2(5x)+\frac1{25}ln[cos(5x)]-\frac12x^2\]

Answer 41

sorry im trying to make sense with what u did to what we are supposed to do. why is the first sec term - x and not 1?

Answer 42

its a little rule i like to refer to as distribution :)

x tan^2(5x)

x (sec^2(5x) - 1)

x sec^2(5x) - x

Answer 43

i see no point in dragging along a -x thru integration by parts when it can integrate on its own

Answer 44

i set u to x right? and du to dx right?

Answer 45

i would, since a poly always goes down to zero eventually

Answer 46

my solution above accords with the last full solution posted by @amistre64

Answer 47

so dv = integral tan^2(5x)dx and v = tan^2(5x). I am using then sec^2 to evaluate... so would by new dv be equal to integral sec^2(5x)-1 dx?

Answer 48

dv = sec^2(5x)
v = tan(5x)/5

Answer 49

tan^2(5x) does not integrate up to tan^2(5x)

since we have to integrate, and we know that tan derives to sec^2

i just see it simpler to integrate sec^2 back up to a tan

Answer 50

yes i agree.. i am getting confused I guess with the substitution.. it seems the computation is not the issue

Answer 51

My Math Lab set dv equal to the integral.. that kind of messed me up.. they drug that x all the way down to the bottom and skipped like 4 steps to get the final answer on the example problem.

Answer 52

x tan^2(5x) is equal to: x sec^2(5x) - x, the -x part is basic enough and can be worked on its own

x sec^2(5x) needs some love :)

u = x v = ? [tan(5x)/5]
du = dx dv = sec^2(5x) dx

uv - int v du

x tan(5x)/5 - int tan(5x)/5 dx

Answer 53

so you leave the 1 out and put it at the end.. so v only ever equals 1/5tan(5x)

Answer 54

distribute the x thru so that its not a -1 on the end, but a -x ....

Answer 55

but when u do u*V wont it be there? and again when u do vdu?

Answer 56

instead of doing

u = x v = tan(5x)/5 - x
du = dx dv = sec^2(5x) - 1

theres no sense to me in dragging that extra part along for the ride

Answer 57

the negative 1 u mean.. so u just integrate it from the beginning and tack it on as x^2/2 at the end?

Answer 58

do you agree that the integration of a sum is the sum of integrations?

\[\int a+b=\int a+\int b\]

Answer 59

yes

Answer 60

do you agree that:\[a(x+y)=ax+ay\]

Answer 61

yes

Answer 62

so rewrite it as integral sec^(5x) - integral 1 dx ?

Answer 63

and, do you agree that:\[x~tan^2(5x)=x~(sec^2(5x)-1)\\~~~~~~~~~~~~~~~~~=x~sec^2(5x)~-~x\]

Answer 64

yep

Answer 65

then we have no -1 on the end, we have a -x

\[\int x~sec^2(5x)~-~x\]
\[\int x~sec^2(5x)~-\int~x\]
do you agree?

Answer 66

yes

Answer 67

then the rest is already set in stone

Answer 68

ok i understand what u are saying.. so let me clarify though... so when i set dv to \[\int\limits \sec^2(18x)-\int\limits 1 dx\]

I do not have to include the int 1dx into v?

Answer 69

I know you said you dont wanna take it for the ride when u can integrate is easily.. I am just trying to understand the rules at which u moved it out to the end.

Answer 70

i see you want to not distribute and jsut take the sec^2-1 along for the ride

u = x v = tan(5x)/5 - x <-- you would have to bring it
du = dx dv = sec^2(5x) - 1 along

Answer 71

\[x(\frac15tan(5x)-x)-\left(\int \frac15tan(5x)-x~dx\right)\]
\[\frac x5tan(5x)-x^2-\left(\int \frac15tan(5x)~dx-\int x~dx\right)\]

Answer 72

!!!!!! I UNDERSTAND WOOO !!!!! ha i wrote it down again with a new problem and now, moving the integral makes sense... well not by math rules.. but...

Answer 73

its doable .. but just extra work

Answer 74

yay!! my daughter is happy that we can leave now :)

Answer 75

sorry :( thanks for helping me for like an hour! my apologies

Answer 76

youre welcome, and good luck :)

Answer 77

thanks.. this calc II class is like 5505% harder than calc I lol