Question

Please help I'm stuck but I have a start find the integral of sinx/1+cos^2x would anyone turn the numerator into sin^2x?

Answer 1

@zepdrix @ganeshie8 @vishweshshrimali5

Answer 2

Denomiator*

Answer 3

try \[u=\cos(x)\]should work immediately

Answer 4

I did u=1+cos^2x probably why I'm having trouble

Answer 5

\[\int \frac{\sin(x)dx}{1+\cos^2(x)}\]right ?

Answer 6

Yes

Answer 7

you get \[-\int \frac{1}{1+u^2}du\]right a way , which should look like a familiar derivative

Answer 8

think of a well know function whose derivative is \[\frac{1}{1+x^2}\]

Answer 9

Inverse tangent? I think

Answer 10

yes

Answer 11

believe it or not, with the minus sign included it is actually the derivative of arccotangent, but no one ever used that

Answer 12

*uses

Answer 13

Yeah, interesting so how would I know when to just use cos instead of cos^2x+1?

Answer 14

practice

Answer 15

plus if you put \[u=1+\cos^2(x)\] then you get \[du=-2\cos(x)\sin(x)\] which is no where in sight

Answer 16

I was going to say when I did it that's probably why I got stuck because I didn't know whAt to do after I found the du= part

Answer 17

Right

Answer 18

it is like learning how to factor
just as tedious, just as boring, and for the most part just as useless, although some people dig it

Answer 19

Lol yes :D

Answer 20

i would not include myself in the group that finds it at all interesting

Answer 21

It's okay, thank you for your help :)

Answer 22

yw

Answer 23

:)