Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx=\alpha {\log }_e|1+\tan x|+\beta {\log }_e|1-\tan x+{\tan }^{2}x|+\gamma {\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C$, when $C$ is constant of integration, then the value of $18(\alpha +\beta +{\gamma }^2)$ is.

Unlocking the Integral

Welcome, fellow JEE aspirant! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions. But as you will soon see, the beauty of mathematics lies in its hidden order.

We are tasked with evaluating the integral $I=\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx$ and matching it to a specific form to find the value of $18(\alpha +\beta +{\gamma }^2)$. Let us embark on this journey together.

The Homogeneous Transformation

When you encounter an integral where the denominator is a sum of powers of $\sin x$ and $\cos x$, your first instinct should be to look for homogeneity. Here, both terms in the denominator are of degree 3. This is a massive hint!

To simplify this, we divide both the numerator and the denominator by ${\cos }^{3}x$. Watch what happens:

In the numerator, we split $\frac{\sin x}{{\cos }^{3}x}$ into $\frac{\sin x}{\cos x}\cdot \frac{1}{{\cos }^{2}x}$, which is simply $\tan x{\sec }^{2}x$. In the denominator, $\frac{{\sin }^{3}x}{{\cos }^{3}x}$ becomes ${\tan }^{3}x$, and $\frac{{\cos }^{3}x}{{\cos }^{3}x}$ becomes $1$.

Our integral now looks like this:

This is the 'Aha!' moment. We have successfully transformed a complex trigonometric expression into a clean, algebraic one.

The Substitution

Now, let us make our lives even easier. We see $\tan x$ and its derivative ${\sec }^{2}x$ sitting right there. Let $t=\tan x$.

Then, $dt={\sec }^{2}xdx$. Substituting these into our integral, we get:

We have completely eliminated the trigonometry! Now, we are dealing with a classic rational function. To integrate this, we must factorize the denominator using the sum of cubes identity: $1+t^3=(1+t)(t^2-t+1)$.

Our integral becomes:

The Art of Partial Fractions

This is where we perform some algebraic surgery. We want to split this fraction into simpler parts. We set:

Why did we choose $B(2t-1)$? Because $2t-1$ is the derivative of $t^2-t+1$. This is a powerful trick that makes the integration step trivial later on.

By equating coefficients, we find $A=-\frac{1}{3}$, $B=\frac{1}{6}$, and $C=\frac{1}{2}$.

The Final Integration

With our constants found, the integral splits into three parts:

The first two integrals are logarithmic. The third requires completing the square in the denominator: $t^2-t+1=(t-\frac{1}{2}{)}^2+\frac{3}{4}$.

This leads us to the ${\tan }^{-1}$ form. After careful calculation, we get:

Comparing this to the target form, we identify $\alpha =-\frac{1}{3}$, $\beta =\frac{1}{6}$, and $\gamma =\frac{1}{\sqrt{3}}$.

Finally, we calculate:

And there it is! A complex-looking problem reduced to a simple integer. Keep practicing these patterns, and you will find that even the most intimidating integrals have a logical, beautiful path to the solution.