[itex]\int sec(x)tan^{2}(x)[/itex] Can it be integrated w/ U-Sub?

  • Thread starter Lebombo
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  • #1
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Homework Statement





Last week, the Professor of the class I'm taking jotted down some even and odd exponent rules of thumb to make life easier when integrating various trig functions. Rules like if you have sinx and cosx and one of the trig functions is odd, then split, use an identity and the U-Sub..etc.

My question is: is this a straight forward integration based on the above mentioned type of rules?

[itex]\int sec(x)tan^{2}(x)[/itex]


If I replace the tan^{2}x with its identity (sec^{2}x - 1) I end up with sec^{3} again with an additional secx.

This looks more tricky than simply being able to use those rules.
 
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  • #2
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I speak strictly for myself.

When integrating
[itex]\int \left[ \sec^3x - \sec x \,\right] dx[/itex]

Surely you know you can separate the terms into
[itex]\int \sec^3x \,dx - \int \sec x \,dx[/itex]

Now you should memorized the integral of secant x, so that's out of the way.

You could also memorize the integral of secant cubed, which is nicely demonstrated here.

Then you just back-substitute into the second equation.

Seems like a relatively easy way to integrate in my opinion.
 
  • #3
vela
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Homework Statement





Last week, the Professor of the class I'm taking jotted down some even and odd exponent rules of thumb to make life easier when integrating various trig functions. Rules like if you have sinx and cosx and one of the trig functions is odd, then split, use an identity and the U-Sub..etc.

My question is: is this a straight forward integration based on the above mentioned type of rules?

[itex]\int sec(x)tan^{2}(x)[/itex]


If I replace the tan^{2}x with its identity (sec^{2}x - 1) I end up with sec^{3} again with an additional secx.

This looks more tricky than simply being able to use those rules.
Those rules only work to a certain point. Eventually you get an integral you have to evaluate using other methods, and the integral of ##\sec^3 x## is one of those. Use integration by parts to calculate that one.

You could also forgo the initial use of a trig substitution and evaluate the original integral using integration by parts.
 
  • #4
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Thank you ReneG and Vela.
 

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