# Another trig-integration question.Watch

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#1
a) Prove the identity 2cosec2x = sec²x/tanx

b)Hence, by using the substitution u=tanx, show that the integral of

cosec2x dx= 1/2 ln |tanx| +c

Thanks!

P.s. has anyone got a link to a list of all the trig identities and/or the standard integrals?
0
14 years ago
#2
(Original post by jumpunderaboat)
a) Prove the identity 2cosec2x = secx/tanx

b)Hence, by using the substitution u=tanx, show that the integral of

cosec2x dx= 1/2 ln |tanx| +c

Thanks!

P.s. has anyone got a link to a list of all the trig identities and/or the standard integrals?
There is something wrong with that identity.
0
#3
(Original post by J.F.N)
There is something wrong with that identity.
fixed!
0
14 years ago
#4
Making an assumption that what you mean is prove that 2cosec2x=(((secx)^2)/tanx)

a) (((secx)^2)/tanx)=(1/(cosxcosx))(cosx/sinx)=1/sinxcosx=1/sinxcosx*1

=>(2/2)(1/sinxcosx)=(2/2sinxcosx)

*2sinxcosx=sin2x=>=(2/sin2x)=2cosec2x

b) INT[cosec2x]dx=(1/2)INT[(((secx)^2)/tanx)]dx=INT[(1/u)(du/dx)*dx]

=>=(1/2)lnu+C

*u=tanx

=>=(1/2)lntanx+C

Newton.
0
14 years ago
#5
(Original post by jumpunderaboat)
a) Prove the identity 2cosec2x = sec²x/tanx

b)Hence, by using the substitution u=tanx, show that the integral of

cosec2x dx= 1/2 ln |tanx| +c

Thanks!

P.s. has anyone got a link to a list of all the trig identities and/or the standard integrals?
2csc2x=2/2sinxcosx=1/sinxcosx=secx.1/sinx
multiply by (1/cos x)/(1/cosx) to get =secx.(1/cosx)/(sinx/cosx)=secx.secx/tanx=sec²x/tanx

b) cosec2x=sec²x/2tanx
u=tan x, du/dx=sec²x -->dx=du/sec²x
--> integrate du/2u --> this gives 1/2ln|u| + c --> 1/2ln|tan x| +c
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