Can You Crack This Challenging Indefinite Integral?

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Discussion Overview

The discussion revolves around solving a challenging indefinite integral. Participants share their attempts and methods for finding the integral, exploring various approaches and tools used in the process.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in solving the integral and seeks assistance, indicating that their known methods are insufficient.
  • Another participant shares a result obtained from Maple, presenting the integral in terms of a numerator and denominator, suggesting
GodsmacK
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Hello, everyone

I've trying to solve this integral but it seems like the methods I know are not enogh to solve it. So I'd be glad if you could give me some trick to get into the answer.
Here it is:

33wtvdw.jpg


Thanks in advance!
 
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Maple knocked it out right away (I have no idea what the details are):

Integral = num/den
where
num =(x*e^x*(tan((1/2)*e^x))^2-x*e^x+2*tan((1/2)*e^x))
den=(1+tan((1/2)*e^x))*e^x

Perhaps having the result in hand will enable you to go back and fill in the gaps.
 
I found it to be e-xsin(ex) - xcos(ex) + c
This was through Wolfram and I would guess integration by parts somehow
 
Hey guys!

I've just solved this thing. In fact, like Charles wrote, it is integration by parts. First you distribute the \sin(e^x) into the parenthesis, then you do the substitution u=e^x. After some steps you shall get something like this:

I=\int \ln(u) \sin(u)du-\int \frac{\sin(u)}{u^2}du

Applying integration by parts:

I=-\cos(u) ln(u)+\int \frac{\cos(u)}{u}du+\frac{\sin(u)}{u}-\int \frac{\cos(u)}{u}du

And there you are... As you see the non-primitive function integral gets cancelled.
 
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Eureka! And substituting back reduces it down. Aren't integrals just a blast?
 

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