Question

How do I integrate (sin^2x-cos^2x)/sinx?

Answer 1

fair...lol

Answer 2

sinx - cosx*cotx

Answer 3

you could also get it in terms of just sin

cos^2 = 1-sin^2

Answer 4

sinx - (1-sin^2x)/sinx = 2sinx -1/sinx

Answer 5

I think you can do it from here

Answer 6

i hav triied it uhh..uh

Answer 7

Ruchi what did you just paste?

Answer 8

i need an indefinite integral... how did you even get to a numerical solution on an indefinite integral?

Answer 9

you have to integrate it

Answer 10

Ignore that, do what dumbcow told you

Answer 11

yeahhhhhhhhhhhpii

Answer 12

yeah im doing the work on that now... ill get back if i need more help from there

Answer 13

zacha.. m i gona get it ahh?

Answer 14

ruchi you told me nothing...

Answer 15

(sinx)^2+(cosx)^2=1
or (cosx)^2=1-(sinx)^2
So:
1-(sinx)^2-(sinx)^2=sin(x)
1-2(sinx)^2=sin(x)
0=2(sinx)^2+sin(x)-1
sinx=[-1+sqrt(1+8)]/4 or [-1-sqrt(1+8)]/4
sinx=[-1+3]/4 or [-1-3]/4
sinx=2/4 or -4/4
sinx=1/2 or -1
If sinx=0.5
x=pi/6+2piK
or x=5pi/6+2piK
If sinx=-1
x=-pi/2+2piK
All the possible solutions are actually
x=pi/6+(2/3)piK
And -pi<x<=pi
The solutions are:
x= -pi/2, pi/6, 5pi/6

Answer 16

that's solving for x if it were an equation... i'm looking for the INDEFINITE INTEGRAL.

Answer 17

so i simplified and got integral(3sinx-cscx) which= ln|cscx+cotx|-2cosx+C

Answer 18

sorry that should be a 2 in front of the sin

Answer 19

integrate (sin^2x-cos^2x)/2sinx
-( cos^2x - sin^2x) = -(1 - 2sin^2x) = 2sin^2x -1
->(sin^2x-cos^2x)/2sinx = sinx - 1/2sinx
Int (sinx - 1/2sinx) dx = -cosx + ln |cscx + cotx| + c

Answer 20

OOps, missing 2 at Ln.
= -cosx + (1/2) ln |cscx + cotx| + c