intergral

**alexandrabel90** · January 9th 2010, 02:37 PM

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))?

**Drexel28** · January 9th 2010, 03:04 PM

Originally Posted by alexandrabel90

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))?

$\int_0^{\ln(\pi)}e^x \sin\left(e^x\right)=\int_1^\pi \sin\left(z\right)dz=-\cos(\pi)+\cos(1)=1+\cos(1)$

**HallsofIvy** · January 10th 2010, 06:04 AM

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 $\int_{ln_0}^{\pi} sin(u)du= \left(cos(u)\right)_0^{ln(\pi)}$ $= cos(ln(\pi))- cos(0)= - cos(\ln(\pi))-1$ .

Well, there are two errors in that. The first, small error, is that is the integral from 0 to $ln(\pi)$ , 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, $\int e^x sin(e^x)dx= \int sin(u)du= -cos(u)+ C$ . Now, change back to the variable x. Since $u= e^x$ , that is the same as $- cos(e^x)+ C$ . At the upper limit, $ln(\pi)$ , that is $-cos(\pi)+ C= 1+ C$ . At the lower limit, 0, that is $- cos(1)+ C$ . Subtracting, $1+ C- (-cos(1)+ C)= 1+ cos(1)$ 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, $u= e^x= e^0= 1$ . When $x= ln(\pi)$ , $u= e^{\ln(\pi)}= \pi$ . Now we have that $\int_{x=0}^{ln(\pi)}e^x sin(e^x)dx= \int_{u=1}^{\pi} sin(u)du$ [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 $\int_{x=0}^{ln(\pi)}e^x sin(e^x)dx= \int_{u=1}^{\pi} sin(u)du$ .

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