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Integration using substitution - HSN forum

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Integration using substitution


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#1 YIC

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Posted 20 January 2006 - 09:49 PM

could someone help me with this question?

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

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

Help would be nice smile.gif

#2 dfx

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Posted 20 January 2006 - 10:40 PM

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

integral.gif (Sinx/Cos power2.gif x) = integral.gif (-1/u power2.gif) du

I think.

#3 YIC

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Posted 20 January 2006 - 10:48 PM

does sec power2.gif x = 1/cos power2.gif x ? That is confusing because the derivative of sec power2.gif x is 1/cos power2.gif x. But on the other hand, secx = 1/cosx so sec power2.gif x must equal 1/cos power2.gif x.

Cheers for the help man biggrin.gif

#4 Steve

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Posted 22 January 2006 - 06:40 PM

QUOTE(YIC @ Jan 20 2006, 11:48 PM)
the derivative of sec power2.gif x is 1/cos power2.gif x

No, it isn't smile.gif



\begin{align*}\frac{d}{dx}\left( \sec^2x\right) &= \frac{d}{dx}\left( (\cos x)^{-2}\right) \\
&= -2.(\cos x)^{-3}.(-\sin x) \\
&= \frac{2\sin x}{(\cos x)^{3}}
\end{align*}
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#5 YIC

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Posted 24 January 2006 - 10:28 PM

Oops my mistake!

I copied something down wrong down in my notes.

#6 Vixus

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Posted 11 February 2006 - 10:18 PM



\begin{align*}\frac{2 sin x}{(cos x)^3}
= 2 \frac{sin x}{cos x} * \frac{1}{(cos x)^2}
= 2 \frac{tan x}{cos^2 x}\end{align*}





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