# Integration by Trigonometric Substitution.

1. Oct 13, 2007

### azatkgz

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

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

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

Last edited: Oct 13, 2007
2. Oct 13, 2007

### 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?

3. Oct 13, 2007

### azatkgz

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

4. Oct 13, 2007

### learningphysics

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

5. Oct 13, 2007

### azatkgz

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

6. Oct 13, 2007

### learningphysics

ok. so this part is wrong

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

7. Oct 13, 2007

### Gib Z

Express sec in terms of tan again.

Last edited: Oct 13, 2007
8. Oct 13, 2007

### azatkgz

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

9. Oct 13, 2007

### 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.

10. Oct 13, 2007

### dextercioby

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