Question

Check my answer(Calculus)

Answer 1

My answer is the one on the bottom.

Answer 2

you need to use the integrating factor method

Answer 3

My answer is wrong?

Answer 4

\[e^( \int\limits \tan \Theta d \theta) = e^(-\ln \cos \theta) = 1/\cos \theta =( the integrating factor) \]

Answer 5

yes it seems to be

Answer 6

I went:
\[(\sin \Theta * dr/d \Theta + \cos \Theta * r)/ \sin \theta = \tan \theta /\sin \theta\] --> \[dr/d \theta + \cot \theta * r = \sec \theta\]

Answer 7

\[r/\cos \theta = \int\limits \tan \theta /\cos \theta d \theta \]

Answer 8

ok so the integrating factor is sin theta

Answer 9

\[r/\sin \theta = \int\limits \tan \theta /\sin \theta d \theta =\int\limits 1/\cos \theta d \theta = \ln|\sec \theta + \tan| + C \]

Answer 10

\[r= (\ln|\sec \theta+ \tan| + C)/\sin \theta\]

Answer 11

and we are done

Answer 12

How did you go from r * cos0 to r/cos0?

Answer 13

where

Answer 14

start from were i have rsin0

Answer 15

The original equation had cos0 *r, and to remove the sine from dr/d0, we divide everything by sin0 right? So then you would get cos0/sin0 * r, which would be cot0 * r?

Answer 16

i made a mistake

Answer 17

u'll get a integrating factor of sinx right

Answer 18

Right

Answer 19

then rsinO= int sinO * tanO dO ok

Answer 20

so from the integrating factor I went:

sin0(dr/d0 + cot0 * r) = tan0
took hte integral
sin0 * r = -ln(cos0) + c

Answer 21

then the int sinO*tanO dO= -Log[Cos[O/2] - Sin[O/2]] + Log[Cos[O/2] + Sin[O/2]] - Sin[O]

Answer 22

Wouldn't you divide to get r?

Answer 23

rsinO= int sinO*tanO dO= -Log[Cos[O/2] - Sin[O/2]] + Log[Cos[O/2] + Sin[O/2]] - Sin[O]

Answer 24

now r = (int sinO*tanO dO= -Log[Cos[O/2] - Sin[O/2]] + Log[Cos[O/2] + Sin[O/2]] - Sin[O])/sinO

Answer 25

sorry its so drummbled

Answer 26

view this http://www.wolframalpha.com/input/?i=(sin%5E2(x))%2Fcos(x)

Answer 27

okay, thanks!