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How do you integrate #int x^2 csc x dx # using integration by parts?

Answer 1

Luke Alvarez

You will need some more advanced mathematics than just integration by parts.

WolframAlpha gives an answer involving complex values functions including the complex valued polylogarithm function.

#int x^2cscx dx = 2ix(Li_2(-e^(ix)) - Li_2(e^(ix))) + 2(Li_3(e^(ix))-Li_3(e^(-ix)))+x^2(ln(1-e^ix)-ln(1+e^(ix)))#

#Li_n(x)# is the polylogarithm function (No. I cannot explain that.)

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Answer 2

Emma Aldridge

To integrate ∫x^2 csc(x) dx using integration by parts, we choose ( u = x^2 ) and ( dv = csc(x) dx ). Then, we find ( du ) and ( v ) as follows:

( du = 2x dx )
( v = -ln|csc(x) + cot(x)| )

Applying the integration by parts formula:

[ \int u dv = uv - \int v du ]

we get:

[ \int x^2 csc(x) dx = -x^2 ln|csc(x) + cot(x)| - \int (-ln|csc(x) + cot(x)|)2x dx ]

Simplify and integrate the remaining term if possible.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.