Integration advice needed (1 Viewer)

leehuan · Jun 24, 2016

How would you tackle integrals of the form

$\int \sec^m{x}\tan^n{x}\,dx$

where m is odd and n is even?

(Within the scope of MX2)

Paradoxica · Jun 24, 2016

Use integration by parts to construct a two-step recurrence relation in both variables.

Alternatively, convert it into the fundamental trigonometric ratios and reduce the cosines and sines.

leehuan · Jun 24, 2016

Paradoxica said:
Use integration by parts to construct a two-step recurrence relation in both variables.

Alternatively, convert it into the fundamental trigonometric ratios and reduce the cosines and sines.

Oh dear. So there isn't a trivial substitution like the two other cases have?

InteGrand · Jun 24, 2016

I'm afraid this case is the most tedious of the bunch.

photastic · Jun 24, 2016

What they mean by equations 1 is the circle identity of secx and tanx, and the derivative of tanx and secx.

leehuan · Jun 24, 2016

photastic said:
What they mean by equations 1 is the circle identity of secx and tanx, and the derivative of tanx and secx.

Hm. What's an example integral where l is even?

InteGrand · Jun 24, 2016

leehuan said:
Hm. What's an example integral where l is even?

$\noindent Say $I = \int \sec ^5 x \tan^ 4 x \text{ d}x$. Then $\tan^4 x = \left(\sec^2 x-1\right)^2$. So $I = \int \sec ^5 x \left(\sec^2 x-1\right)^2\text{ d}x$. Expanding this via the binomial theorem will give us some terms with an even (and some with odd) power of $\sec x$.$

leehuan · Jun 24, 2016

InteGrand said:
$\noindent Say $I = \int \sec ^5 x \tan^ 4 x \text{ d}x$. Then $\tan^4 x = \left(\sec^2 x-1\right)^2$. So $I = \int \sec ^5 x \left(\sec^2 x-1\right)^2\text{ d}x$. Expanding this via the binomial theorem will give us some terms with an even (and some with odd) power of $\sec x$.$

Wait... Did I make a stupid in my WolframAlpha?

http://www.wolframalpha.com/input/?i=expand+sec^5x(sec^2x-1)^2

InteGrand · Jun 24, 2016

leehuan said:
Wait... Did I make a stupid in my WolframAlpha?

http://www.wolframalpha.com/input/?i=expand+sec^5x(sec^2x-1)^2

Oh sorry lol, I was thinking of a sec rather than sec^2 being the thing expanded.

$\noindent If we have $\sec^m x\tan ^n x$, where $m$ is an odd positive integer and $n$ is an even positive integer, then $n =2k$ for some positive integer $k$ and so $\tan ^n x = \left(\tan ^2 x\right)^k = \left(\sec^2 x -1\right)^k$. Powers of $\sec x$ in this expansion will clearly all be even and then multiplying it with the $\sec ^m x$, they'll all become odd as $m$ is odd. So in the end, all the secant powers will be odd powers, which can be tediously computed using a secant reduction formula method for instance.$

leehuan · Jun 24, 2016

InteGrand said:
Oh sorry lol, I was thinking of a sec rather than sec^2 being the thing expanded.

$\noindent If we have $\sec^m x\tan ^n x$, where $m$ is an odd positive integer and $n$ is an even positive integer, then $n =2k$ for some positive integer $k$ and so $\tan ^n x = \left(\tan ^2 x\right)^k = \left(\sec^2 x -1\right)^k$. Powers of $\sec x$ in this expansion will all be even and then multiplying it with the $\sec ^m x$, they'll all become odd as $m$ is odd. So in the end, all the secant powers will be odd powers, which can be tediously computed using a secant reduction formula method for instance.$

Ouch. Figures, my suspicions were right. (Might need a tiny revision in those notes @ photastic)

Thank goodness we have reduction formulae I guess...

InteGrand · Jun 24, 2016

leehuan said:
Ouch. Figures, my suspicions were right. (Might need a tiny revision in those notes @ photastic)

Thank goodness we have reduction formulae I guess...

Yeah, you may want to pre-derive that formula before setting out to tackle an integral like this so that you have it handy if you need it.

Paradoxica · Jun 24, 2016

As an aside, what about the ever so disgusting tangent half angle substitution?

Is that a viable method?

InteGrand · Jun 24, 2016

It's probably horrifyingly tedious most of the time. But in principle it'd work.

photastic · Jun 24, 2016

leehuan said:
Ouch. Figures, my suspicions were right. (Might need a tiny revision in those notes @ photastic)

Thank goodness we have reduction formulae I guess...

Lol they're from MATH1251

leehuan · Jun 24, 2016

photastic said:
Lol they're from MATH1251

HAHAHAHAHA

Sent from my iPhone using Tapatalk

braintic · Jun 25, 2016

leehuan said:
How would you tackle integrals of the form

$\int \sec^m{x}\tan^n{x}\,dx$

where m is odd and n is even?

(Within the scope of MX2)

Actually the Ext 2 syllabus specifically states that these questions are NOT part of the course:

"Recurrence relations such as ....... which involve more than one integer parameter, are excluded."

InteGrand · Jun 25, 2016

braintic said:
Actually the Ext 2 syllabus specifically states that these questions are NOT part of the course:

"Recurrence relations such as ....... which involve more than one integer parameter, are excluded."

Maybe leehuan meant when the m and n are particular numerical values rather than parameters? E.g. finding something like ∫sec³ x tan² x dx.

Paradoxica · Jun 25, 2016

braintic said:
Actually the Ext 2 syllabus specifically states that these questions are NOT part of the course:

"Recurrence relations such as ....... which involve more than one integer parameter, are excluded."

I see no reason as to why double parameter integrals should not be part of the HSC.

braintic · Jun 26, 2016

InteGrand said:
Maybe leehuan meant when the m and n are particular numerical values rather than parameters? E.g. finding something like ∫sec³ x tan² x dx.

Actually he probably did. I didn't read the post properly and missed "of the form".

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