Integration by Trigonometric Substitution.

[azatkgz]

I'm not sure about answer.It looks very strange.

Homework Statement

\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}

The Attempt at a Solution

for u=lnx-->u'=1/x
\int \frac{du}{\sqrt{1+u^2}}
substituting u=tan\theta

=\int \frac{d\theta}{cos\theta}=ln|sec\theta+tan\theta|

\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}=ln|\sqrt{-1}|

 

[learningphysics]

Your formula looks right but your final results isn't. Didn't you actually use u = tan(theta) ?

what limits do you get for theta?

 

[azatkgz]

I just put to the
ln|lnx+\sqrt{lnx-1}|

 

[learningphysics]

I just put to the
ln|lnx+\sqrt{lnx-1}|

Did you use u = tan(theta) or u = sec(theta) ?

 

[azatkgz]

Sorry i typed wrongly.I used u=tan(theta)

 

[learningphysics]

Sorry i typed wrongly.I used u=tan(theta)

ok. so this part is wrong

ln|lnx+\sqrt{lnx-1}|

fix your substitution.

 

[Gib Z]

Express sec in terms of tan again.

 

[azatkgz]

but if I change limits
\int_{0}^{\frac{\pi}{4}}sec\theta d\theta=ln2

 

[Gib Z]

I don't know what you just did, but continue as you were before, you had the right anti derivative: ln |tan O + sec O|, but you didn't replace the original variable back in properly for the sec O.

 

[dextercioby]

Just use a "sinh" substitution when you're left only with the sqrt in the denominator and you're done.