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krisshP

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I need help on q3e. I don't understand how

dv/dx = 2secxsecxtanx

means that

v=sec^2x

Please explain in depth.

Thanks

ghostwalker

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(Original post by krisshP)
[A
I don't understand how

dv/dx = 2secxsecxtanx

means that

v=sec^2x

Please explain in depth.

Thanks

Use a substitution of u=sec x when integrating "2secxsecxtanx" wrt x.

Alternativley, differentiate sec^2x and see what you get.

It doesn't require integration by parts.

krisshP

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(Original post by ghostwalker)
Use a substitution of u=sec x when integrating "2secxsecxtanx" wrt x.

Alternativley, differentiate sec^2x and see what you get.

It doesn't require integration by parts.

Sorry, but so far I haven't done integration by substitution, but I have done integration by parts. Maybe I should just reattempt the question after doing integration by substitution? Annoying thing is this question was actually in the chapter of integration by parts

.

Thanks anyway.

ghostwalker

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(Original post by krisshP)
Sorry, but so far I haven't done integration by substitution, but I have done integration by parts. Maybe I should just reattempt the question after doing integration by substitution? Annoying thing is this question was actually in the chapter of integration by parts

.

Thanks anyway.

OK. You can do it by IBP.

Split as sec d(sec)/dx

krisshP

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(Original post by ghostwalker)
OK. You can do it by IBP.

Split as sec d(sec)/dx

Thanks for that. I'll give it a shot later and see what happens

ghostwalker

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(Original post by krisshP)
Thanks for that. I'll give it a shot later and see what happens

You should get something like

Spoiler:

Show

$I = 2\sec^2x - I$

Where "I" is your original integral.

and re-arrange.

Khallil

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(Original post by krisshP)

I need help on q3e. I don't understand how

dv/dx = 2secxsecxtanx

means that

v=sec^2x

Please explain in depth.

Thanks

$\dfrac{dv}{dx} = 2\sec^2 x \tan x \not\implies v(x) = \sec^2 x$

By noticing that

$\displaystyle \int 2\sec^2 x \tan x\ dx$

is of the form

$\displaystyle \int [f(x)]^{n} f'(x)\ dx = \dfrac{[f(x)]^{n+1}}{n+1}$ ,

you should be able to deduce the value of

flou_fboco2

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(Original post by krisshP)
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I need help on q3e. I don't understand how

dv/dx = 2secxsecxtanx

means that

v=sec^2x

Please explain in depth.

Thanks

If i were you I would intergrate the components seperatly (not sure if this is right though)

krisshP

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Report Thread starter 5 years ago

(Original post by Khallil)

By noticing that

$\displaystyle \int 2\sec^2 x \tan x\ dx$

is of the form

$\displaystyle \int [f(x)]^{n} f'(x)\ dx = \dfrac{[f(x)]^{n+1}}{n+1}$ ,

you should be able to deduce the value of

It worked

.

Thanks

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