anonymous
  • anonymous
integral from 0 to pi/15 of xtan^2(5x) dx please help.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
u=x and du=dx v=1/5tan(5x)-x I am setting up the problem now. uv-integral vdu
anonymous
  • anonymous
\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-\frac{ 1 }{ 5 }(\frac{ 1 }{ 5 }\ln |\sec(5x)|-x)dx\]
anonymous
  • anonymous
\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-x)dx\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
tan^2 = 1-sec^2 which might make life a little easier
anonymous
  • anonymous
i used that for substitution before i started integrating
anonymous
  • anonymous
i set dv = integral sec^2(5x)-1 dx
anonymous
  • anonymous
\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-x)dx| x=\pi/15 and x=0\]
amistre64
  • 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
  • amistre64
of course the 5 has to fit somplace :)
anonymous
  • 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
  • anonymous
ok does this look right to start evaluating at now?
amistre64
  • 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
  • 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
  • anonymous
after simplifying I will end up getting -pi(pi5sqrt(3)-pi/something)
anonymous
  • anonymous
\[\left[ \frac{ \pi }{75 }\sqrt{3}-\frac{ \pi^2 }{ 225 } \right]\]
amistre64
  • 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
  • amistre64
i see, tan^2 = sec^2 - 1
amistre64
  • 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
  • anonymous
the answer i need to get to is \[\frac{ \pi \sqrt{3} }{ 75 }-\frac{ 1 }{ 25 }\ln2-\frac{ \pi^2 }{ 450 }\]
amistre64
  • amistre64
5*15 = 75 2*15^2 = 450 ln(1/2) = -ln(2)
anonymous
  • anonymous
it is the pi*sqrt(3)/75 that i cannot figure out
amistre64
  • amistre64
pi * tan(pi/3) / (5*15) since tan(pi/3) = sqrt(3) ....
anonymous
  • anonymous
\[\left[ \frac{ \pi }{75 }\sqrt{3}-\frac{ \pi^2 }{ 225 } \right]\] what about the last term? how does that just disappear.
amistre64
  • amistre64
\[ \frac x5~tan(5x)\] \[ \frac{\pi}{5*15}~tan(60^o)\] \[ \frac{\pi\sqrt3}{75}\]
anonymous
  • anonymous
\[=x(\frac{ 1 }{ 5 }\tan(5x)-x)-(\frac{ 1 }{ 25 }\ln |\sec(5x)|-\frac{ x^2 }{ 2 }\]
anonymous
  • anonymous
coming from this.. do you not distribute to the last term?
anonymous
  • anonymous
or should the -x not be in there? for uv it is x * 1/5 tan (5x) -x
anonymous
  • anonymous
\[=\pi/15(\frac{ 1 }{ 5 }\tan(5\pi/15)-\pi/15)\] maybe this isnt right?
amistre64
  • 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
  • amistre64
at x=0 that all goes to zero, so its a focus on x=pi/15
anonymous
  • anonymous
isnt the formula uv-integral udv ?
anonymous
  • anonymous
thus if u = x and v = 1/5tan(5x)-x ... man i feel dumb lol
amistre64
  • 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
  • amistre64
the stand alone x can be integrated on its one
anonymous
  • anonymous
\[\frac x5\tan(5x) +\frac1{25} \ln[\cos(5x)] - \frac{1}{2}x^2\]
anonymous
  • anonymous
in the first part of this... where is the other x?
amistre64
  • amistre64
that x^2/2 on the end is the result of integrates -x
anonymous
  • anonymous
\[\frac x5\tan(5x) \]
anonymous
  • anonymous
isnt that part of the vdu? isnt vdu 1/5tan(5x)-x?
IrishBoy123
  • IrishBoy123
amistre64
  • 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
  • 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
  • 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
  • amistre64
i see no point in dragging along a -x thru integration by parts when it can integrate on its own
anonymous
  • anonymous
i set u to x right? and du to dx right?
amistre64
  • amistre64
i would, since a poly always goes down to zero eventually
IrishBoy123
  • IrishBoy123
my solution above accords with the last full solution posted by @amistre64
anonymous
  • 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
  • amistre64
dv = sec^2(5x) v = tan(5x)/5
amistre64
  • 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
  • anonymous
yes i agree.. i am getting confused I guess with the substitution.. it seems the computation is not the issue
anonymous
  • 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
  • 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
  • anonymous
so you leave the 1 out and put it at the end.. so v only ever equals 1/5tan(5x)
amistre64
  • amistre64
distribute the x thru so that its not a -1 on the end, but a -x ....
anonymous
  • anonymous
but when u do u*V wont it be there? and again when u do vdu?
amistre64
  • 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
  • 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
  • amistre64
do you agree that the integration of a sum is the sum of integrations? \[\int a+b=\int a+\int b\]
anonymous
  • anonymous
yes
amistre64
  • amistre64
do you agree that:\[a(x+y)=ax+ay\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
so rewrite it as integral sec^(5x) - integral 1 dx ?
amistre64
  • amistre64
and, do you agree that:\[x~tan^2(5x)=x~(sec^2(5x)-1)\\~~~~~~~~~~~~~~~~~=x~sec^2(5x)~-~x\]
anonymous
  • anonymous
yep
amistre64
  • 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
  • anonymous
yes
amistre64
  • amistre64
then the rest is already set in stone
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • amistre64
its doable .. but just extra work
amistre64
  • amistre64
yay!! my daughter is happy that we can leave now :)
anonymous
  • anonymous
sorry :( thanks for helping me for like an hour! my apologies
amistre64
  • amistre64
youre welcome, and good luck :)
anonymous
  • anonymous
thanks.. this calc II class is like 5505% harder than calc I lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.