MAT244-2018S > Final Exam

FE-P4

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Tim Mengzhe Geng:

--- Quote from: Syed_Hasnain on April 13, 2018, 02:37:46 AM ---The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c

--- End quote ---
$1/(costsint)=2/sin(2t)=2csc(2t)$
And by the integration of 2csc(2t),
we get
$-\ln|csc(2t)+cot(2t)|+c$

Tim Mengzhe Geng:

--- Quote from: Syed_Hasnain on April 13, 2018, 02:37:46 AM ---The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c

--- End quote ---
I'm so sorry that you're also correct  and the two result are the same since
$\tan\frac{x}{2}=\frac{\sin(x)}{1+cos(x)}$
and therefore
$-\ln|csc(2x)+cot(2x)|=\ln|tan(x)|$

Victor Ivrii:
Solution is complete, but the answer must be written.

Also a simpler form $\ln(\tan(t))$ is preferable.

Finally \sin, \cos, and so on must be escaped by \ to provide upright letters and a proper spacing

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