Question

help plzzz ....calculus ....

Answer 1

u need a question lol

Answer 2

\[\int\limits \frac{ x+sinx }{ 1+cosx }dx\]

Answer 3

@amistre64 @ganeshie8 @dan815

Answer 4

might want to split it up

Answer 5

the right side is logable, the the left side by parts

Answer 6

\[\int\limits \frac{ x+sinx }{ 1+cosx }dx\]
\[\int\limits~ x\frac{1}{ 1+cosx }+\frac{ sinx }{ 1+cosx }~dx\]

Answer 7

@amistre64 by parts it takes long i am not getting ... any other method .....ans is
x tan (x/2)

Answer 8

im wondering of there is some nifty little multiply by one format that could be used then, or an add zero trick

Answer 9

i recall the first time a saw the sec(x) solution .... i was sooo mad at it lol

Answer 10

well this my first sum in this excersise which is taking soo long .by parts for the above its tooo confusing ,,,,

Answer 11

\[\int sec(x)~dx\]
\[\int sec(x)\frac{sec(x)+tan(x)}{sec(x)+tan(x)}~dx\]
\[\int \frac{sec^2(x)+sec(x)tan(x)}{sec(x)+tan(x)}~dx\]
\[\int \frac{sec^2(x)+sec(x)tan(x)}{tan(x)+sec(x)}~dx=\ln(tan(x)+sec(x))\]

Answer 12

this is tooo diffrent .....@@

Answer 13

logarithmic integral

Answer 14

@ amistre64 i think i confused u ...:-((....

Answer 15

\[\int\limits~ \frac{x}{ 1+cosx }+\frac{ sinx }{ 1+cosx }~dx\]
\[\int\limits~ \frac{x}{ 1+cosx }dx-\ln[1+cos(x)]\]
\[\int\limits~ \frac{x}{ 1+cosx }dx\]

i agree, this is a pain :/

Answer 16

plain??

Answer 17

p-a-i-n; splitting it up dint make it easier, or at least by parts on this doesnt make it any easier

Answer 18

hmm .... too confused with this ......

Answer 19

anyone solve the by parts ... i would be great help
????

Answer 20

can u do the derivative of x tan x

Answer 21

or x tan(x/2)
and work backwards