Question

intergral of square root of[1+(cosx)^2]

Answer 1

what the................

Answer 2

This? \(\Large \int \sqrt{1+(cosx)^2}~dx\)?

Answer 3

yes

Answer 4

you can do a change of variable
and take X=cos(x) and instead of integrating over IR you will integrate over [-1,1]

Answer 5

it's just an idea let's see if it will work

Answer 6

let's replace dx first using X
dX=-sin(x)dx
let's replace again sin(x) by an expression using X
so we know that (sin(x))^2+(cos(x))^2=1

Answer 7

we obtain then
\[\sin(x)=\sqrt{1-\cos^2(x)} \]
we replace cos(x) by X
then \[\sin(x)=\sqrt{1-X}\]

Answer 8

sorry it's \[\sin(x)=\sqrt{1-X^2}\]

Answer 9

wow thanks. I was stuck for the past 30min.]

Answer 10

you're welcome except I don't know where it's goinbg to lead it's just an idea ;)

Answer 11

I am sorry for my stupid question. I don't get!! since the original problem is \(\Large \int \sqrt{1+(cosx)^2}~dx\), not \(\Large \int \sqrt{1-(cosx)^2}~dx\)
but after manipulate the expression , you get the latter one, how to convert to the original one?

Answer 12

I am computing the length of a curve where the original fn is y=sinx on the interval [0;pi]

Answer 13

@Loser66 that's just dX=-sinxdx=-root(1-X^2)dx
so you have dx=dX/-root(1-X^2)

Answer 14

according to what he did applying change of variable