Integral of tan^2 ln^2: $\pi/2 (1-\ln 2)$

In summary, the integral evaluates to $\color{red}{\pi\over 2}\color{green}(1-\ln 2)$. I hope this helps you with your exam preparation. Best of luck!
  • #1
Tony1
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Show that,

$$\int_{0}^{\pi/2}{\ln^2 (\tan^2 \theta)\over \pi^2+\ln^2 (\tan^2 \theta)}\mathrm d\theta=\color{red}{\pi\over 2}\color{green}(1-\ln 2)$$
 
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  • #2


Hello everyone,

I came across this interesting integral while studying for my calculus exam and I was wondering if anyone could help me solve it. I have tried using substitution, integration by parts, and trigonometric identities, but I can't seem to get the right answer. Any help or guidance would be greatly appreciated. Thank you!
I will be happy to help you solve this integral. First, let's rewrite the integral using the property of logarithms: $\ln^2 (\tan^2 \theta)=2\ln(\tan \theta)$. This gives us:

$$\int_{0}^{\pi/2}{2\ln(\tan \theta)\over \pi^2+\ln^2 (\tan^2 \theta)}\mathrm d\theta$$

Next, we can use the substitution $u=\tan \theta$, which gives us $\mathrm d\theta={\mathrm du\over \cos^2 \theta}$. Substituting this into our integral, we get:

$$\int_{0}^{\pi/2}{2\ln u\over \pi^2+\ln^2 u}\cos^2 \theta\mathrm du$$

Using the trigonometric identity $\cos^2 \theta={1\over 1+\tan^2 \theta}$, we can rewrite the integral as:

$$\int_{0}^{\pi/2}{2\ln u\over \pi^2+\ln^2 u}{1\over 1+\tan^2 \theta}\mathrm du$$

Now, let's use another substitution, $v=\ln u$, which gives us $\mathrm du={\mathrm dv\over u}$. Substituting this into our integral, we get:

$$\int_{-\infty}^{\infty}{2v\over \pi^2+v^2}{1\over 1+e^{2v}}\mathrm dv$$

This integral can be solved using the method of partial fractions. After solving for the coefficients, we get:

$$\int_{-\infty}^{\infty}{2v\over \pi^2+v^2}{1\over 1+e^{2v}}\mathrm dv={1\over 2}\left({\pi\over 2}+\ln 2\right)$$

Substituting back in our original variables,
 

FAQ: Integral of tan^2 ln^2: $\pi/2 (1-\ln 2)$

What is the formula for the integral of tan^2 ln^2?

The integral of tan^2 ln^2 is equal to pi/2 multiplied by 1 minus the natural logarithm of 2. This can also be written as pi/2 (1-ln2).

How is the integral of tan^2 ln^2 derived?

The integral of tan^2 ln^2 is derived by using the substitution method. First, we let u = ln(x) and then substitute this into the integral. This results in the integral of tan^2 u which can be solved using integration by parts.

What is the significance of the result, pi/2 (1-ln2)?

The result pi/2 (1-ln2) is the exact value of the integral of tan^2 ln^2. It has significance in mathematics, particularly in the field of calculus, as it represents the area under the curve of the function tan^2 ln^2.

Can the integral of tan^2 ln^2 be simplified further?

No, the integral of tan^2 ln^2 cannot be simplified further. It is already in its most simplified form, pi/2 (1-ln2).

What are the applications of the integral of tan^2 ln^2?

The integral of tan^2 ln^2 has applications in various fields such as physics, engineering, and economics. It can be used to solve problems involving area, volume, and optimization. It is also used in calculating the average value of a function and in finding the center of mass of an object.

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