Question

Looking for the integral of the following....
sinx/(1+sin^2x)

Answer 1

integral sinx*1/1+sin^2x

take u = sinx

Answer 2

but du/dx = cosx dx

Answer 3

Hmm that causes some trouble with the cosine that shows up D: yah

Answer 4

du/dx = cosx hence dx = du/cosx so we cant substitute

Answer 5

i'm missing something?

Answer 6

ok ok ok ok :D
i will always think of converting it to complex number :3
it looks more neat hehehe

Answer 7

missing what ?

Answer 8

converrrrtttt yeahhhhhhh

Answer 9

hoy wanna work on complex ?

Answer 10

u will only have one branch cut

Answer 11

Chad. If we differentiate your answer we get sinxcosx/(1+sin^2x)

Answer 12

ohh a bit cool , continue wih trig identety
u will got something like (k sin 2 x / m ( cos 2 x))

Answer 13

the numerator and the denomominator divided by cos^2 (x).
the integral bocomes :
int ( sec(x)tan(x)/(sec^2 (x) + tan^2 (x)) dx
= int sec(x)tan(x)/(2sec^2 (x) - 1)

u = sec(x)
du = sec(x)tan(x) dx

= int du/(2u^2 - 1)

Answer 14

lavoshneeey

Answer 15

i think partial fraction is next step ....

Answer 16

we would use it in first place instead

Answer 17

but to me that only sound need to be solved with complex assumtion

Answer 18

hmm but if u make it bounded u will only get an answer of real number, trust me -.-
do you know how complex work ?

@chad123 it looks hmm something wrong , u assumed first sin x = x then where is dx ?

Answer 19

yeah i stuffed up what im trying to do is let sinx = x

Answer 20

we will then have integral x/x^2+1 dx

Answer 21

but then u have to make
cos x dx = dx

Answer 22

so it will be

x cos x / 1+ x^2 dx

Answer 23

u dint get it right ?
its wrong

Answer 24

1/(2u^2 - 1) = A/(sqrt(2)u+1) + B/(sqrt(2)u-1)
i got A = -1/2 and B = 1/2

Answer 25

Your on the right track tanjung but i get 1/(2u^2-1) = 1/2(u+1/sqrt2)(u-1/sqrt2)

Answer 26

then i get 1/2(A/(u+1/sqrt2) + B/(u-1/sqrt2)

Answer 27

and i get A= -1/sqrt2 B= 1/sqrt2