Animated Solution for Mathematics - Indefinite Integration: Evaluate $\int \frac{(\cos 2x{)}^{1/2}}{\sin x}dx$

Conquering the Radical

Welcome, fellow traveler on the path to JEE mastery. Today, we face an integral that might look like a tangled mess of trigonometric functions, but I promise you, it is a beautiful puzzle waiting to be solved.

We are looking at the integral:

The square root is the first thing that catches our eye, and it is usually the first thing that causes panic. But in the world of JEE, panic is just a lack of strategy.

The Unification

Whenever you see a double angle like $2x$ mixed with a single angle $x$, your first instinct should be to unify them. We have $\cos 2x$ in the numerator, so let us use the classic double angle identity: $\cos 2x={\cos }^{2}x-{\sin }^{2}x$.

Substituting this into our integral, we get:

Now, look at that $\sin x$ in the denominator. If we bring it inside the square root, it becomes ${\sin }^{2}x$. This is a neat trick that allows us to split the fraction inside the root:

Our complex integral has beautifully simplified to:

The Radical Removal

We still have a square root to deal with. Think about trigonometric identities; we know that ${\sec }^2\theta -1={\tan }^2\theta$. This perfectly matches our structure.

Let us make a strategic substitution. We will let $\cot x=\sec \theta$. By doing this, the expression inside the root becomes $\sqrt{{\sec }^2\theta -1}$, which is simply $\tan \theta$. The root is gone!

The Differential Dance

Since we changed our variable to $\theta$, we must also find our new differential, $d\theta$. Differentiating both sides of $\cot x=\sec \theta$, we get:

Isolating $dx$, we have $dx=-\frac{\sec \theta \tan \theta }{{\csc }^{2}x}d\theta$. Remembering that ${\csc }^{2}x=1+{\cot }^{2}x=1+{\sec }^2\theta$, our differential becomes:

Now, let us piece everything back together. Multiplying the $\tan \theta$ from the root and the $\tan \theta$ from $dx$ gives us ${\tan }^2\theta$. Our integral becomes:

The Rational Transformation

When faced with a mess of secants and tangents, the best strategy is often to convert everything back to sines and cosines. After some algebraic manipulation, we arrive at:

By substituting ${\cos }^2\theta =1-{\sin }^2\theta$, we get:

This is a favorite concept of JEE. We substitute $u=\sin \theta$, then $du=\cos \theta d\theta$. Our integral transforms into a rational function:

Using partial fractions, we decompose this into:

The Final Reveal

Now, we integrate these standard forms. The integral of $\frac{1}{a^2-u^2}$ is $\frac{1}{2a}\log \left|\frac{a+u}{a-u}\right|$. Applying this, we return to our original variable $x$.

The final, magnificent answer is:

Take a moment to appreciate the journey from that initial integral to this result. You have conquered the beast!