may i know why the answer for integrate exp (x) sin ( exp x) from x= ln (pi) to x= 0 is not

1- cos (ln (pi))?

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- Jan 9th 2010, 02:37 PMalexandrabel90intergral
may i know why the answer for integrate exp (x) sin ( exp x) from x= ln (pi) to x= 0 is not

1- cos (ln (pi))? - Jan 9th 2010, 03:04 PMDrexel28
- Jan 10th 2010, 06:04 AMHallsofIvy
My first reaction was to wonder why you think it

**should**be! But I think I see. I suspect you made the substitution u= e^{x} so that the integral became $\displaystyle \int_{ln_0}^{\pi} sin(u)du= \left(cos(u)\right)_0^{ln(\pi)}$$\displaystyle = cos(ln(\pi))- cos(0)= - cos(\ln(\pi))-1$.

Well, there are**two**errors in that. The first, small error, is that is the integral**from 0 to $\displaystyle ln(\pi)$**, not the other way around. I suspect you didn't mean it that way, anyway.

The second, more important, error is that the original limits of integration, ln(x) and 0, apply to the original variable, x, not u! There are two ways of handling that correctly.

The first is to ignore the limits of integration until the end. As an indefinite integral, $\displaystyle \int e^x sin(e^x)dx= \int sin(u)du= -cos(u)+ C$. Now, change back to the variable x. Since $\displaystyle u= e^x$, that is the same as $\displaystyle - cos(e^x)+ C$. At the upper limit, $\displaystyle ln(\pi)$, that is $\displaystyle -cos(\pi)+ C= 1+ C$. At the lower limit, 0, that is $\displaystyle - cos(1)+ C$. Subtracting, $\displaystyle 1+ C- (-cos(1)+ C)= 1+ cos(1)$ which is what Drexel28 got.

The other, better way, to do this is to change the limits of integration as you go. When x= 0, $\displaystyle u= e^x= e^0= 1$. When $\displaystyle x= ln(\pi)$, $\displaystyle u= e^{\ln(\pi)}= \pi$. Now we have that $\displaystyle \int_{x=0}^{ln(\pi)}e^x sin(e^x)dx= \int_{u=1}^{\pi} sin(u)du$[tex]= \left(-cos(u)\right)_{u=1}^\pi= (-cos(\pi))- (-cos(1))= 1+ cos(1) again.

That's one reason I advocate writing the limits of integration in the form "x= ..." as I did in $\displaystyle \int_{x=0}^{ln(\pi)}e^x sin(e^x)dx= \int_{u=1}^{\pi} sin(u)du$.