Question

Solve y'=(y^2+y*tan(x)+tan^2 x)/(sin^2 x) explicitly.

Answer 1

\[y' = \frac{y^2+y*\tan{x}+\tan^2{x}}{\sin^2{x}}\]
Try \(y = -\tan{x}\tan{f(x)} \) which would imply that y' was of the form:
\[-\tan{x}*\sec^2{f(x)}*f'(x) - \sec^2{x}*\tan{f(x)}\]

Answer 2

@dumbcow @hartnn

Answer 3

What should my substitution be? Is it y=tan(x)*t?

Answer 4

not sure...hold on let try a couple of things

Answer 5

alright looks like that will work, it makes it separable

\[y = (\tan x) t\]
y' = (sec^2)t + (tan) t'

... a lot of simplifying

---> dt/(t^2+1) = dx/sincos

Answer 6

See? I got t'=(t^2+1)/(cos(x)*sin(x)) and that's where I got stucked.

Answer 7

ahh change t' to dt/dx

Answer 8

I know t'=dt/dx, so after I got dt/(t^2+1)=dx/(sin(x)*cos(x)), I know that I need to integrate, but how do I integrate the right side dx/(sin(x)*cos(x))?

Answer 9

2 sin x cos x = sin 2x

1/sin2x = ... ?

Answer 10

I'm confused. I know that 2 sinx cos x=sin 2x and that's the identity, but we have sin x cos x on the denominator. How do we integrate that?

Answer 11

1/ sin x cos x = 2/sin 2x = 2 cosec x

Answer 12

2 cosec 2x i mean

Answer 13

\[\sin(x)\cos(x)=\frac{1}{2} \cdot 2 \sin(x)\cos(x)=\frac{1}{2}\sin(2x) \\ \text{ so } \frac{1}{\sin(x)\cos(x)}=\frac{1}{\frac{1}{2}\sin(2x)}=2\csc(2x)\]

Answer 14

and nice sub cow

Answer 15

So how do we integrate 2csc2x?

Answer 16

\[\int\limits_{}^{}\csc(x) dx=\int\limits_{}^{}\csc(x) \cdot \frac{\csc(x)+\cot(x)}{\csc(x)+\cot(x)} dx \\ \text{ Let } u=\csc(x)+\cot(x) \\ du =-\csc(x)(\cot(x)+\csc(x)) dx \\ \int\limits_{}^{}\frac{-du}{u}=-\ln|u|+C\\=-\ln|\csc(x)+\cot(x)|+C\]

Answer 17

use a substitution

Answer 18

another approach is change 1/sincos = sec^2/ tan
u = tan
du = sec^2 dx

Answer 19

i thought u can use the standard result of integrating cosec x

Answer 20

Wait a minute, guys, I'll come back.

Answer 21

you can @hartnn
I was just trying to show how to get the standard result
and just in case he wondered why

Answer 22

So I got tan^-1 t=ln abs(tan x)+C after integrating. How do I solve for t?

Answer 23

@Idealist10 , im just gonna post rest of solution for you to verify later

\[\int\limits \frac{dt}{t^2 +1} = \int\limits \frac{\sec^2 x}{\tan x} dx\]
\[\tan^{-1} (t) = \ln (\tan x) +C\]
\[t = \tan[ \ln(\tan x) +C]\]
sub back in for y
\[y = (\tan x) \tan[ \ln(\tan x) +C]\]

Answer 24

Thank you so much for the big help!

Answer 25

yw
ive never seen so many tangents in 1 problem :)

Answer 26

i saw more than 5 tangents in one problem....but the question belonged to geometry and circles :P

Answer 27

More than 5 tangents in one problem? That's insane.

Answer 28

but there were 2 circles 2 :P