Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{\cos x}{{\sin }^{3}x(1+{\sin }^{6}x{)}^{2/3}}dx=f(x)(1+{\sin }^{6}x{)}^{1/3}+c$, where $c$ is a constant of integration, then $\lambda f(\frac{\pi }{3})$ is equal to :

Welcome, my fellow traveler on the path to JEE excellence. Today, we are not just solving a problem; we are peeling back the layers of a complex integral to reveal the elegant structure hidden beneath.

Look at the expression on your screen:

It looks daunting, doesn't it? But in the world of JEE, complexity is often just a mask for a beautiful, simple path.

The first thing we must do is observe. Notice the $\cos x$ in the numerator? That is not a coincidence; it is a beacon.

We know that the derivative of $\sin x$ is $\cos x$. This is our golden ticket. Let us perform our first substitution: $t=\sin x$.

Differentiating both sides gives us $\cos xdx=dt$. Now, watch as the trigonometric fog lifts:

Now, we face the core challenge. We have $t^3$ in the denominator and a bracket $(1+t^6{)}^{2/3}$.

The trick, a classic JEE maneuver, is to force a substitution by extracting the highest power of $t$ from inside the bracket. We factor out $t^6$ from $(1+t^6{)}^{2/3}$.

When $t^6$ leaves the bracket, it carries the exponent $2/3$ with it, becoming $(t^6{)}^{2/3}=t^4$. Inside the bracket, we are left with $(1+t^{-6})$.

So, our denominator becomes $t^3\cdot t^4(1+t^{-6}{)}^{2/3}$, which simplifies beautifully to $t^7(1+t^{-6}{)}^{2/3}$. Our integral is now:

Now, look at the term inside the bracket: $1+t^{-6}$. Its derivative is $-6t^{-7}$.

Look at our integral again. We have $dt$ in the numerator and $t^7$ in the denominator, which is exactly $t^{-7}dt$. Let us make our second substitution: $u=1+t^{-6}$.

Then $du=-6t^{-7}dt$, which means $\frac{dt}{t^7}=-\frac{1}{6}du$. Substituting this into our integral, we get:

This is a standard power rule integral. Integrating $u^{-2/3}$ gives us $3u^{1/3}$. Multiplying by $-1/6$, we get:

We are almost there. We must return to our original variable $x$. First, substitute $u=1+t^{-6}$ back:

Writing $t^{-6}$ as $1/t^6$ and simplifying, we get:

Finally, replace $t$ with $\sin x$:

Comparing this with the given form $f(x)(1+{\sin }^{6}x{)}^{1/3}+c$, we identify $f(x)=-\frac{1}{2{\sin }^{2}x}$. To find the final answer, we calculate $3f(\frac{\pi }{3})$.

Since $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$, then ${\sin }^2(\frac{\pi }{3})=\frac{3}{4}$. Thus:

Multiplying by $3$, we get the final result: -2.