Question

I have gotten myself all confused now...

(sin x)/(sin x)^2 +1....I can cancel the sin x?

brain is saying yes other part of brain is saying are you sure? other part of brain is saying you will regret it....not sure who to listen to

Answer 1

If you cancel it, you are assuming that it is not 0. Do you really want to make that assumption?

Answer 2

Also is the problem:\[\frac{\sin x}{\sin^2x+1} or \frac{\sin x}{\sin^2x}+1\]

Answer 3

the one is in the denominator so example 1

Answer 4

Then absolutely not. It cannot be cancelled. Only factors may be cancelled. NOT addends.

Answer 5

So what exactly are you supposed to be doing with this problem?

Answer 6

I wrote myself into a corner trying to answer a question from purplemouse way down the page....I think my attempt hit a deadend...sadly lol

Answer 7

Good luck

Answer 8

∫1/((sinx)2+1)∗(sinx)/1

seems like there should be a way out of here...

Answer 9

Where is Turing or Satellite when you need them

Answer 10

hmm that should be an sinx^2 in the denominator...didnt paste well

Answer 11

\[\frac{\sin(x)}{\sin^2(x)}=\sin(x)\] for sure

Answer 12

but not
\[\frac{\sin(x)}{\sin^2(x)+1}\]

Answer 13

see there I should listen to my brain...just not sure which part

Answer 14

\[\int\limits_{?}^{?}\frac{sinx}{\sin^2x+1}dx\]

Answer 15

Isn't that the problem?

Answer 16

indeed

Answer 17

hmm maybe
\[\frac{\sin(x)}{\sin^2(x)+1}=\frac{\sin(x)}{2-\cos^(x)}\] then a u - sub

Answer 18

\[\frac{\sin(x)}{\sin^2(x)+1}=\frac{\sin(x)}{2-\cos^2(x)}\]

Answer 19

u=cosx and du=-sin(x)dx

Answer 20

\[\int\limits_{?}^{?}1/(u^2 + 2) du\]