## anonymous 5 years ago Help with integrals

1. anonymous

what do you need help with?

2. anonymous

$\int\limits_{}^{}\frac{dx}{\sin^2x}$

3. anonymous

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

4. anonymous

hold on let me write it out ok?

5. anonymous

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

6. anonymous

sure. thanks

7. anonymous

So how is it?

8. anonymous

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

9. anonymous

Nah.

10. anonymous

-1/tan x sorry wrong answer.

11. anonymous

How did you get it?

12. anonymous

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.

13. anonymous

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.

14. anonymous

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.

15. anonymous

$\text{Let } t = \frac{\tan{x}}{2} \implies \frac{\mathbb{d}x}{\mathbb{d}t} = \frac{2}{1+t^2}$ 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.

16. anonymous

It did work, after all :D

17. anonymous

gj Newton :P that's impressive.

18. anonymous

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 :(

19. anonymous

THAT;S A GREAT HELP. THANKS!

20. anonymous

Ugh, the sub is $t = \tan \frac{x}{2}$ So many typos :/ But you're welcome.

21. anonymous

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