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#1
Report 16 years ago
#1
Hi,

Would really appreciate if anyone could help me with this problem:

Integrate

y=(x)sec(x^2)

Clueless, please help! Thanks!
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you need to do integration by parts. let x be v and sec^2 du. That should work

peace
flamma
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sorry should have added you need do do substitution 1st let t = x^2

peace
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Flakker
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#4
Report 16 years ago
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right, y=xsec^2(x).
integral[xsec^2(x)] is required.


ok. I'd do integration by parts, forgive me if I screw this up, I have p3 on tues so it'll cost me!

let u = x
and let dv/dx=sec^2(x)

1. du/dx= 1

2. v = integral[dv/dx]=int[sec^2(x)]= tan (x)

3.now, assuming you're doing edexcel maths, turn to page 8 of the formulae sheet, look at the bottom expression in the 'integration' section.

You will see:


integral[u(dv/dx)dx] = uv - integral[v(du/dx)dx]

thus we have:

integral[xsec^2(x)]= x*tan(x)- int[tan(x) * 1]

Note that the integral of tan(x) is ln|sec(x)| + C

integral[xsec^2(x)]= xtan(x)- ln|sec(x)| + C



At least, I hope so! Or else Im in trouble.
Chris
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Eru Iluvatar
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#5
Report 16 years ago
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Well im only on P1 at the moment, but that sounds about right to me.
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#6
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Chris it's meant to be integrate

x*(sec(X^2)), but i think what you integrated is right. Good luck.
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