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Integration using substitution - HSN forum

# Integration using substitution

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### #1YIC

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Posted 20 January 2006 - 09:49 PM

could someone help me with this question?

sinx sec[^2]x dx , using the substitution: u = cosx

so far I have got to du/u * du , but this seems like a load of bollocks.

Help would be nice

### #2dfx

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Posted 20 January 2006 - 10:40 PM

u = Cosx so du/dx = -Sinx so du = -Sinxdx

(Sinx/Cos x) = (-1/u ) du

I think.

### #3YIC

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Posted 20 January 2006 - 10:48 PM

does sec x = 1/cos x ? That is confusing because the derivative of sec x is 1/cos x. But on the other hand, secx = 1/cosx so sec x must equal 1/cos x.

Cheers for the help man

### #4Steve

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Posted 22 January 2006 - 06:40 PM

QUOTE(YIC @ Jan 20 2006, 11:48 PM)
the derivative of sec x is 1/cos x

No, it isn't

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### #5YIC

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Posted 24 January 2006 - 10:28 PM

Oops my mistake!

I copied something down wrong down in my notes.

### #6Vixus

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Posted 11 February 2006 - 10:18 PM

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