Challenge integral (1 Viewer)

Luukas.2

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There is a problem with the answer here. The function is positive throughout its domain. Yet, using the result presented:


The result as written does work for the domain , though.
 

Luukas.2

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We need to factorise the denominator:



The integrand can now be decomposed using partial fractions:




 

Luukas.2

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also what application did u use to do ur working
I typed it in the box for posting replies.

Note that the original solution posted:


does works on some domains - for example, on .

However, for the domain , the integral is


This tells me that there is an issue with signs or in making the assumption that when, in fact and is only when .

Note, for example, that I did not make the substitution as it would follow that ... instead, I substituted and avoided a potential problem. Far too many papers include substitutions that raise problems that are then ignored, yielding correct answers but from technically-flawed working.
 

Luukas.2

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Now for a follow up question: hence or otherwise integrate sqrt(tanx):)

The partial fractions result will now yield both inverse tan and log terms.

Alternatively, focus on


and then subtract the result from the integral found earlier in this thread.
 

SB257426

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I typed it in the box for posting replies.

Note that the original solution posted:


does works on some domains - for example, on .

However, for the domain , the integral is


This tells me that there is an issue with signs or in making the assumption that when, in fact and is only when .

Note, for example, that I did not make the substitution as it would follow that ... instead, I substituted and avoided a potential problem. Far too many papers include substitutions that raise problems that are then ignored, yielding correct answers but from technically-flawed working.
Brilliant answer btw
 

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We need to factorise the denominator:



The integrand can now be decomposed using partial fractions:




Good answer but there's an easier way than doing partial fraction decomposition, so I thought I might share it:



The logic is basically trying to find a way to cancel the nominator with a substitution so


then making the substitution,

which should lead to the same solution but simplified with the tan inverses combined
 
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The partial fractions result will now yield both inverse tan and log terms.

Alternatively, focus on


and then subtract the result from the integral found earlier in this thread.
nice, i was intending for it to be done by sqrt(tanx) = 1/2 (sqrttanx + sqrtcotx + sqrttanx - sqrtcotx) tho
 

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