- #1

- 4

- 1

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:

Thanks in advance!

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- Thread starter GodsmacK
- Start date

- #1

- 4

- 1

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:

Thanks in advance!

- #2

OldEngr63

Gold Member

- 732

- 51

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.

- #3

- 31

- 5

This was through Wolfram and I would guess integration by parts somehow

- #4

- 4

- 1

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

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

Applying integration by parts:

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

And there you are... As you see the non-primitive function integral gets cancelled.

- #5

- 31

- 5

Eureka! And substituting back reduces it down. Aren't integrals just a blast?

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