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
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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.
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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.
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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
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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
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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.
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Khallil
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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
\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 v(x)
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flou_fboco2
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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)
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krisshP
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(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 v(x)
It worked.

Thanks
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