- #1

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Hi,

find value of

[tex]

\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx

[/tex]

I first used subsitution u = lnx, that gave me:

[tex]

\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

then letting

[tex]

I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

integrating by parts twice, gave me:

[tex]

I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du

[/tex]

[tex]

I = - e^{u} \cos u + e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

therefore

[tex]

I = - e^{u} \cos u + e^{u} \cos u + I

[/tex]

which of course cancels to zero which is not very useful

Any ideas on finding the value of this integral, or have i made an error in my working?

Many Thanks

## Homework Statement

find value of

[tex]

\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx

[/tex]

## Homework Equations

## The Attempt at a Solution

I first used subsitution u = lnx, that gave me:

[tex]

\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

then letting

[tex]

I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

integrating by parts twice, gave me:

[tex]

I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du

[/tex]

[tex]

I = - e^{u} \cos u + e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du

[/tex]

therefore

[tex]

I = - e^{u} \cos u + e^{u} \cos u + I

[/tex]

which of course cancels to zero which is not very useful

Any ideas on finding the value of this integral, or have i made an error in my working?

Many Thanks

Last edited: