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December 31st, 2011, 10:24 PM   #1
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integral of sin2x/2sincos^2x

I'm studying integral calculus right now..

and my teacher gave as assignments..

I really don't have idea to answer this

integral of sin2x/2sincos^2x

but I try to answer this.. and I got

I rewrite it to:
2sinxcosx/2sincos^2x

then divide them and I got 1/cosx

integral of sec x = ln |sec x + tan x| + c

ln |sec x + tan x| + c this was my final answer.. is this correct?

Happy new year and thank you in advance.
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December 31st, 2011, 10:38 PM   #2
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Re: integral of sin2x/2sincos^2x

I assume you mean:



Let:

and we have:



You are correct.
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December 31st, 2011, 10:53 PM   #3
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Re: integral of sin2x/2sincos^2x

Ohh my, am I really correct?

Wow... Thank you very much.... as in super thank you

She also gave this homework...

integral of dx / 1-cosx I let u = sin x du =x dx=du

integral of dx / u = dx/ sinx = 1/sinx = integral of csc x

then my final answer was -ln|csc x + cot x | + c


Am I also correct here?

Thank you again...
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December 31st, 2011, 10:58 PM   #4
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Re: integral of sin2x/2sincos^2x

Quote:
Originally Posted by MarkFL
I assume you mean:



Let:

and we have:



You are correct.

Sir the denominator is only 2sincos^2x...

is it ok to divide with 2sinxcosx... and get 1 because there is x before cosx in numerator

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December 31st, 2011, 11:21 PM   #5
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Re: integral of sin2x/2sincos^2x

We are given:



Observing that we may write:

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January 1st, 2012, 12:05 AM   #6
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Re: integral of sin2x/2sincos^2x

You're so good...

thank you again..
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January 1st, 2012, 05:34 PM   #7
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Quote:
Originally Posted by nephi39
. . . only 2sincos^2x...
That was presumably a typo and should have been 2sin x cosx.

Using cos x = 1 - 2sin(x/2),



The substitution u = sin x (for which du = cos x dx) wouldn't have helped.
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