Integration problem

• March 28th 2010, 07:28 PM
temaire
Integration problem
Use appropriate substitution and than a trigonometric substitution and evaluate the integral.
$\int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}$

My attempt

$\int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}$

$ln y = tan\theta$
$y = cos^{2}\theta$
$dy = -2cos\theta sin\theta d\theta$

$= -2\int_{1}^{e}\frac{cos\theta sin\theta d\theta}{cos^{2}\theta\sqrt{1 + tan^{2}\theta}}$

$= -2\int_{1}^{e}\frac{sin\theta d\theta}{cos\theta sec\theta}$

$= -2\int_{1}^{e}sin\theta d\theta$

How do I proceed from here? I think I have to change the limits of integration in terms of $\theta$ instead of $y$.
• March 28th 2010, 07:40 PM
sa-ri-ga-ma
Substitute ln(y) = t.
1/y*dy = dt.
Substitute the limits of y and find the limit of t.
• March 28th 2010, 07:44 PM
Sudharaka
Quote:

Originally Posted by temaire
Use appropriate substitution and than a trigonometric substitution and evaluate the integral.
$\int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}$

My attempt

$\int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}$

$ln y = tan\theta$
$y = cos^{2}\theta$
$dy = -2cos\theta sin\theta d\theta$

$= -2\int_{1}^{e}\frac{cos\theta sin\theta d\theta}{cos^{2}\theta\sqrt{1 + tan^{2}\theta}}$

$= -2\int_{1}^{e}\frac{sin\theta d\theta}{cos\theta sec\theta}$

$= -2\int_{1}^{e}sin\theta d\theta$

How do I proceed from here? I think I have to change the limits of integration in terms of $\theta$ instead of $y$.

Dear temaire,

I think you have substituted wrong.

Substitute, $\theta=tan^{-1}(lny)$

$ln y = tan\theta$

$\frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta$

$\frac{dy}{y}=sec^{2}{\theta}~d{\theta}$

Also, you have to change the limits...

When, $y=1\Rightarrow{\theta=0}$

When, $y=e\Rightarrow{\theta=\frac{\pi}{4}}$

Then your integration would be, $\int_{0}^{\frac{\pi}{4}}{\frac{sec^{2}\theta~d\the ta}{\sqrt{1+tan^{2}\theta}}}$
Can you do it form here???
• March 28th 2010, 07:45 PM
temaire
Quote:

Originally Posted by sa-ri-ga-ma
Substitute ln(y) = t.
1/y*dy = dt.
Substitute the limits of y and find the limit of t.

Shouldn't I use $y = cos^{2}\theta$ to find the limits, by substituting 1 and e into y to get the two new limits in terms of $\theta$
• March 28th 2010, 07:52 PM
temaire
Quote:

Originally Posted by Sudharaka
Dear temaire,

I think you have substituted wrong.

Substitute, $\theta=tan^{-1}(lny)$

$ln y = tan\theta$

$\frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta$

$\frac{dy}{y}=sec^{2}{\theta}~d{\theta}$

Also, you have to change the limits...

When, $y=1\Rightarrow{\theta=0}$

When, $y=e\Rightarrow{\theta=\frac{\pi}{4}}$

Then your integration would be, $\int_{0}^{\frac{\pi}{4}}{\frac{sec^{2}\theta~d\the ta}{\sqrt{1+tan^{2}\theta}}}$
Can you do it form here???

But if $\frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta$, isn't

$y = cos^{2}\theta$?

Therefore, $dy = -2cos\theta sin\theta d\theta$
• March 28th 2010, 07:56 PM
Sudharaka
Quote:

Originally Posted by temaire
But if $\frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta$, isn't

$y = cos^{2}\theta$?

Therefore, $dy = -2cos\theta sin\theta d\theta$

Dear temaire,

You have to use the substitution you use to find the limits. I think you got it confused since your differentiation was wrong. You cannot obtain $y = cos^{2}\theta$. Please recheck how you have differentiated.