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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- amistre64

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

- anonymous

i used that for substitution before i started integrating

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

of course the 5 has to fit somplace :)

- anonymous

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

- anonymous

ok does this look right to start evaluating at now?

- amistre64

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

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

5*15 = 75
2*15^2 = 450
ln(1/2) = -ln(2)

- anonymous

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

- amistre64

pi * tan(pi/3) / (5*15)
since tan(pi/3) = sqrt(3) ....

- anonymous

\[\left[ \frac{ \pi }{75 }\sqrt{3}-\frac{ \pi^2 }{ 225 } \right]\]
what about the last term? how does that just disappear.

- amistre64

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

\[=\pi/15(\frac{ 1 }{ 5 }\tan(5\pi/15)-\pi/15)\]
maybe this isnt right?

- amistre64

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

- amistre64

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

- anonymous

isnt the formula uv-integral udv ?

- anonymous

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

- amistre64

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

- amistre64

the stand alone x can be integrated on its one

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- IrishBoy123

##### 1 Attachment

- amistre64

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

- anonymous

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?

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- IrishBoy123

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

- anonymous

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?

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

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.

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

do you agree that the integration of a sum is the sum of integrations?
\[\int a+b=\int a+\int b\]

- anonymous

yes

- amistre64

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

- anonymous

yes

- anonymous

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

- amistre64

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

- anonymous

yep

- amistre64

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?

- anonymous

yes

- amistre64

then the rest is already set in stone

- anonymous

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?

- anonymous

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.

- amistre64

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

- amistre64

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

- anonymous

!!!!!! 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...

- amistre64

its doable .. but just extra work

- amistre64

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

- anonymous

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

- amistre64

youre welcome, and good luck :)

- anonymous

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

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