Integration by substitution (1 Viewer)

Timothy.Siu · Apr 28, 2009

lol is that the fastest way

Trebla · Apr 29, 2009

Unfortunately, this question for some reason required a trigonometric substitution. The natural way to go about is to use partial fractions. However, it may not necessarily be shorter...

lolokay · Apr 29, 2009

from a few lines in:

$\int \frac{sin^2x - cos^2x}{sin^3xcosx}dx\\ \int \frac{1}{sinxcosx} - \frac{cosx}{sin^3x}dx\\ \int \frac{sin^2 + cos^2x}{sinxcosx} - cosxsin^{-3}x dx\\ \int \frac{sinx}{cosx} + \frac{cosx}{sinx} - cosxsin^{-3}x{sinx}dx\\ = -ln|cosx| + ln|sinx| + \frac{1}{2}sin^{-2}x\\ = ln|tanx| + \frac{1}{2}csc^2x$

then sub in values

shaon0 · Apr 29, 2009

lolokay said:
from a few lines in:

$\int \frac{sin^2x - cos^2x}{sin^3xcosx}dx\\ \int \frac{1}{sinxcosx} - \frac{cosx}{sin^3x}dx\\ \int \frac{sin^2 + cos^2x}{sinxcosx} - cosxsin^{-3}x dx\\ \int \frac{sinx}{cosx} + \frac{cosx}{sinx} - cosxsin^{-3}x{sinx}dx\\ = -ln|cosx| + ln|sinx| + \frac{1}{2}sin^{-2}x\\ = ln|tanx| + \frac{1}{2}csc^2x$

then sub in values

yeah you're right. I think Trebla maybe wrong.

lolokay · Apr 29, 2009

they give the same thing

GUSSSSSSSSSSSSS · Apr 30, 2009

but arent the answers are the same as (cosecx)^2 and (cotx)^2 only differ by a constant =S

but who can be bothered checkin thru trebla's solution =P (i admire you for writing that trebla)

shaon0 · Apr 30, 2009

lolokay said:
they give the same thing

yeah, sorry.

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Integration by substitution (1 Viewer)

Timothy.Siu

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Trebla

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lolokay

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shaon0

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