[Maths] Proving~~

withoutwax · Professor Posts: 101 Join date: 2008-08-02

Exam's over..but we never let our brain stops..

Just found a few interesting question..try it out~~

Prove

{integral} 1/sinx dx=ln{tan[1/2 x]} + C

and

{integral} 1/cosx dx=ln{[cos x]/[1-sin x]} + C

Happy working it~~Whahaha bounce

withoutwax · Professor Posts: 101 Join date: 2008-08-02

No one is willing to giv a damn-try on this??

mia · Posts: 63 Join date: 2008-08-01

Hmm...

{integral} 1/sinx dx
I got is ln tanx...
cos differentiate sinx is cos x and integrate 1/sinx would be ln sinx...
haha.... so eventually is (ln sinx)/(cos x) and thus ln tanx
XD.... I know it must be wrong.... but thats how I see it...

for {integral} 1/cosx dx
I thought will get (-ln cos x)/(sin x) ?
LOL...
cos differentiate cosx is -sin x.... how dya get 1-sinx?? confused

Death · Subject: Re: [Maths] Proving~~ Wed Sep 17, 2008 11:31 am

mia wrote:

Hmm...

{integral} 1/sinx dx
I got is ln tanx...
cos differentiate sinx is cos x and integrate 1/sinx would be ln sinx...
haha.... so eventually is (ln sinx)/(cos x) and thus ln tanx
XD.... I know it must be wrong.... but thats how I see it...

for {integral} 1/cosx dx
I thought will get (-ln cos x)/(sin x) ?
LOL...
cos differentiate cosx is -sin x.... how dya get 1-sinx?? confused

I think we has to use trigonometric identity,sin^2 x + cos^2 x = 1,or the sin x = 2t / 1+t^2 or something..

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withoutwax · Professor Posts: 101 Join date: 2008-08-02

Yup..you're rite Death..

Advice to Mia: y don't you diffrentiate ur ans bak..see if u get the ans anot?

angie · Posts: 4 Join date: 2008-09-16

wakaka~~ let t= tan 1/2 x first~~ hehe~~ then start integrate the 1/sin x.... hehe Surprised

withoutwax · Professor Posts: 101 Join date: 2008-08-02

wow..so geng?? try integrate the other one too..

flamenix · Posts: 27 Join date: 2008-08-05

I had done it.. A good question..