Question

Help with integrals

Answer 1

what do you need help with?

Answer 2

\[\int\limits_{}^{}\frac{dx}{\sin^2x}\]

Answer 3

I got stuck here :\[2\int\limits_{}^{}\frac{dx}{1-\cos(2x)}\]

Answer 4

hold on let me write it out ok?

Answer 5

but btw the 1-cos(2x) use trig identitites.

Answer 6

sure. thanks

Answer 7

So how is it?

Answer 8

\[\int\limits 1\div \sin x ^{2} dx = x / sinx ^{2}\]

Answer 9

Nah.

Answer 10

-1/tan x sorry wrong answer.

Answer 11

How did you get it?

Answer 12

It's normally assumed from the fact
\[\frac{\mathbb{d}}{\mathbb{d}x} \cot x = -cosec^2x\]
.... but if you REALLY want to prove it, try the sub
\[t = \tan \frac{x}{2} \]

Haven't tried it, but it almost always works. May be some easy way I'm missing.

Answer 13

Note, proving the differentiation version, if you deem that sufficient, is far easier. But if you did not know the result, the sub I guess would be OK.

Answer 14

moolean, i'd stick with what hes saying, i think he understand integrals better then i do. im nto perfect at it yet, i was just trying to help, but he seems to rlly know what he's talking about.

Answer 15

\[\text{Let } t = \frac{\tan{x}}{2} \implies \frac{\mathbb{d}x}{\mathbb{d}t} = \frac{2}{1+t^2} \]
It quickly follows that:
\[\sin x = \frac{2t}{1+t^2}\]
\[\int \frac{1}{sin^2x} \mathbb{d}x = \int \left(\frac{1+t^2}{2t}\right)^2\cdot \frac{2}{1+t^2} \mathbb{d}t = \int \left(\frac{1+t^2}{2t^2}\right)\cdot \mathbb{d}t \]
\[\int \left(\frac{1+t^2}{2t^2}\right)^2\cdot \mathbb{d}t = \frac{t^2 - 1}{2t} \]

Which, from double angle formulae, is equal to the required result.

Answer 16

It did work, after all :D

Answer 17

gj Newton :P that's impressive.

Answer 18

Thanks - I needed to practice it anyway.

NOTE there should be no squared on the fraction on the last line. some dodgy copy/pasting from above there :(

Answer 19

THAT;S A GREAT HELP. THANKS!

Answer 20

Ugh, the sub is \[ t = \tan \frac{x}{2} \]

So many typos :/

But you're welcome.

Answer 21

geez i should say thank you too, u taught me something and i was only trying to help :P