Animated Solution for Mathematics - Definite Integration: The integral ${\int }_{\pi /6}^{\pi /3}{\sec }^{2/3}x{\csc }^{4/3}xdx$ equal to :

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of fractional powers. We are looking at the integral $I={\int }_{\pi /6}^{\pi /3}{\sec }^{2/3}x{\csc }^{4/3}xdx$.

When you see expressions like this, it is natural to feel a moment of hesitation. But remember: in calculus, complexity is often just a mask. Our job is to peel back that mask.

The Sin-Cos Foundation

Whenever you encounter secant and cosecant in an integral, your first instinct should be to return to the basics. These functions are merely the reciprocals of cosine and sine.

We know that $\sec x=\frac{1}{\cos x}$ and $\csc x=\frac{1}{\sin x}$. Substituting these into our integral, we get:

Suddenly, the problem feels grounded. We are no longer dealing with abstract secants; we are dealing with the fundamental building blocks of trigonometry.

The Golden Strategy

Now, look at the denominator: ${\cos }^{2/3}x{\sin }^{4/3}x$. We have a product of two trigonometric functions with fractional powers. This is a classic pattern in JEE Advanced problems.

We want to create a $\tan x$ term because we know that the derivative of $\tan x$ is ${\sec }^{2}x$. If we can get a ${\sec }^{2}x$ in the numerator, the integral becomes trivial.

To create $\tan x$, we need to divide $\sin x$ by $\cos x$. Since we have ${\sin }^{4/3}x$, we need to divide it by ${\cos }^{4/3}x$. To keep the equation balanced, we must multiply the numerator by the same term:

Combining the cosine terms in the denominator, we add the exponents: $\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2$. So, the denominator becomes ${\cos }^{2}x\cdot {\tan }^{4/3}x$.

Our integral now looks like this:

Since $\frac{1}{{\cos }^{2}x}$ is simply ${\sec }^{2}x$, and $\frac{{\cos }^{4/3}x}{{\cos }^{2}x}={\cos }^{-2/3}x$, we can simplify the expression. However, a more direct path is recognizing that $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$ and the remaining $\cos$ terms cancel out to leave us with:

The Elegant Substitution

This is the moment of triumph. We have $\tan x$ in the denominator and its derivative, ${\sec }^{2}x$, sitting right there in the numerator. Let us set $t=\tan x$. Then $dt={\sec }^{2}xdx$.

But do not forget the limits! We must convert our $x$-limits to $t$-limits: - For $x=\frac{\pi }{6}$, $t=\tan (\frac{\pi }{6})=\frac{1}{\sqrt{3}}=3^{-1/2}$. - For $x=\frac{\pi }{3}$, $t=\tan (\frac{\pi }{3})=\sqrt{3}=3^{1/2}$.

Our integral is now a simple power rule problem:

Applying the power rule $\int t^{n}dt=\frac{t^{n+1}}{n+1}$, we get:

The Final Calculation

Now, we just plug in the limits. Upper limit: $-3(3^{1/2}{)}^{-1/3}=-3\cdot 3^{-1/6}$. Lower limit: $-3(3^{-1/2}{)}^{-1/3}=-3\cdot 3^{1/6}$.

Subtracting the lower limit from the upper limit gives us:

Using the laws of exponents ($3^1\cdot 3^{1/6}=3^{7/6}$ and $3^1\cdot 3^{-1/6}=3^{5/6}$), we arrive at our final answer:

See how the complexity dissolved? By trusting the process and using the right identities, we turned a terrifying integral into a simple algebraic expression. Keep this confidence with you—every problem has a path to simplicity if you look hard enough.