Animated Solution for Mathematics - Indefinite Integration: Find the indefinite integral $\int \cos 2\theta \ln \left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)d\theta$

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of logarithms and trigonometric ratios.

You see an integral like $\int \cos 2\theta \ln \left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)d\theta$ and your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, intimidation is just a mask; our job is to peel back that mask and reveal the elegant structure underneath.

The Power of Simplification

Never rush into a calculation. The most common mistake students make is diving headfirst into Integration by Parts without cleaning the house first.

Look at that argument inside the logarithm: $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$. It is begging to be simplified.

If we divide both the numerator and the denominator by $\cos \theta$, we get $\frac{1+\tan \theta }{1-\tan \theta }$. This is the expansion of $\tan (A+B)$ where $A=\frac{\pi }{4}$ and $B=\theta$.

Suddenly, our terrifying integral transforms into something much more manageable: See? The monster has already lost its teeth.

The Dance of Integration by Parts

Now that we have a clean expression, we apply the ILATE rule. We have a logarithmic function and a trigonometric function, so logarithms are the priority for our $u$.

We set $u=\ln \tan \left(\theta +\frac{\pi }{4}\right)$ and $dv=\cos 2\theta d\theta$.

Differentiating $u$ requires the chain rule. We get:

Converting these to sine and cosine yields:

If we multiply the numerator and denominator by $2$, the denominator becomes the double angle identity $\sin (2\theta +\frac{\pi }{2})$, which is simply $\cos 2\theta$. Thus, $du=2\sec 2\theta d\theta$. It is beautiful, isn't it? The complexity just evaporates.

The Final Assembly

Now we integrate $dv$ to get $v=\frac{\sin 2\theta }{2}$. We plug everything into the Integration by Parts formula: $\int udv=uv-\int vdu$.

Look at that second integral. The $2$ in the numerator and the $2$ in the denominator cancel out perfectly. We are left with $\int \tan 2\theta d\theta$, which is $\frac{1}{2}\ln |\sec 2\theta |$.

The Elegance of the Result

Putting it all together, we arrive at our final answer:

This problem teaches us a vital lesson: complexity is often just a layer of dust covering a simple, elegant truth. When you face a problem that looks impossible, don't fight it with brute force.

Use your identities, simplify the expressions, and trust the process. You have the tools; you just need to be patient enough to use them. Keep practicing, keep questioning, and keep falling in love with the logic of mathematics.