Let n ≥ 2 be a natural number and 0

Q: Why do we factor out $\sin^n \theta$ specifically? Why not just $\sin \theta$?

Great question! When you have a term like $(\sin^n \theta - \sin \theta)^{1/n}$, factoring out $\sin^n \theta$ allows us to use the property $(ab)^k = a^k b^k$. Since $(\sin^n \theta)^{1/n} = \sin \theta$, this effectively 'pulls' a $\sin \theta$ out of the radical, which then cancels with the $\sin \theta$ in the denominator. It is a classic JEE technique to simplify expressions involving roots or fractional powers.

Q: I got stuck on the derivative of $t = 1 - \sin^{-(n-1)} \theta$. How do I handle the negative exponent?

Don't panic! Treat it as a standard chain rule problem. The derivative of $1$ is $0$. For $-\sin^{-(n-1)} \theta$, the power $-(n-1)$ comes down, multiplying the existing negative sign to make it positive $(n-1)$. Then, you subtract $1$ from the exponent, giving you $-\sin^{-(n-1)-1} \theta = -\sin^{-n} \theta$. Finally, multiply by the derivative of $\sin \theta$, which is $\cos \theta$. It all cleans up beautifully!

Q: Is there a way to solve this without substitution?

While you could potentially use trigonometric identities to manipulate the expression, it would be incredibly tedious and prone to errors. The beauty of this problem lies in recognizing the derivative of the inner function. Once you simplify the integrand, the substitution $t = 1 - \sin^{-(n-1)} \theta$ makes the integral trivial. Always look for that 'hidden' derivative!

Question

Let $n \ge 2$ be a natural number and $0 < 	heta < \pi/2$. Then $\int \frac{(\sin^n 	heta - \sin 	heta)^{1/n} \cos 	heta}{\sin^{n+1} 	heta} d	heta$ is equal to : (Where C is a constant of integration)

Options:

(A) $\frac{n}{n^2 - 1} (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$

(B) $\frac{n}{n^2 + 1} (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$

(C) $\frac{n}{n^2 - 1} (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$

(D) $\frac{n}{n^2 - 1} (1 + \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe! Today, we are going to dismantle a problem that, at first glance, might make your heart skip a beat. We are looking at the integral:

$$I = \int \frac{(\sin^n 	heta - \sin 	heta)^{\frac{1}{n}} \cos 	heta}{\sin^{n+1} 	heta} d	heta$$

It looks like a chaotic mess of powers and trigonometric functions, doesn't it? But in the world of JEE Advanced, intimidation is just a mask for elegance. Let us peel back the layers.

### The Power of Factoring

When you see a term like $(\sin^n 	heta - \sin 	heta)^{\frac{1}{n}}$, your intuition should immediately scream 'factorization!' We want to simplify the inside of that bracket. By pulling out the highest power, $\sin^n 	heta$, we transform the expression:

$$(\sin^n 	heta - \sin 	heta)^{\frac{1}{n}} = [\sin^n 	heta (1 - \frac{\sin 	heta}{\sin^n 	heta})]^{\frac{1}{n}}$$

Now, look at that inner fraction: $\frac{\sin 	heta}{\sin^n 	heta}$ is simply $\frac{1}{\sin^{n-1} 	heta}$. Our expression becomes:

$$[\sin^n 	heta (1 - \frac{1}{\sin^{n-1} 	heta})]^{\frac{1}{n}}$$

By distributing the power $\frac{1}{n}$, we get:

$$(\sin^n 	heta)^{\frac{1}{n}} \cdot (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{1}{n}} = \sin 	heta (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{1}{n}}$$

We have successfully extracted a $\sin 	heta$ from the radical!

### The Simplification Dance

Now, let us place this back into our original integral. The $\sin 	heta$ we just extracted will dance with the denominator $\sin^{n+1} 	heta$:

$$I = \int \frac{\sin 	heta (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{1}{n}} \cos 	heta}{\sin^{n+1} 	heta} d	heta$$

Canceling $\sin 	heta$ from the numerator and denominator leaves us with:

$$I = \int (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{1}{n}} \frac{\cos 	heta}{\sin^n 	heta} d	heta$$

Do you see it? The expression $\frac{\cos 	heta}{\sin^n 	heta} d	heta$ is practically begging to be part of a substitution.

### The Substitution

Let us define $t = 1 - \frac{1}{\sin^{n-1} 	heta}$. To make the differentiation crystal clear, write it as $t = 1 - (\sin 	heta)^{-(n-1)}$.

Now, differentiate with respect to $	heta$:

$$\frac{dt}{d	heta} = 0 - [-(n-1) (\sin 	heta)^{-n} \cos 	heta]$$

Which simplifies to:

$$dt = (n-1) \frac{\cos 	heta}{\sin^n 	heta} d	heta$$

Rearranging this, we find that $\frac{\cos 	heta}{\sin^n 	heta} d	heta = \frac{dt}{n-1}$. The integral is now a simple power rule problem:

$$I = \int t^{\frac{1}{n}} \frac{dt}{n-1} = \frac{1}{n-1} \int t^{\frac{1}{n}} dt$$

### Final Calculation

Integrating $t^{\frac{1}{n}}$ gives us $\frac{t^{\frac{1}{n} + 1}}{\frac{1}{n} + 1}$, which is $\frac{t^{\frac{n+1}{n}}}{\frac{n+1}{n}}$. Multiplying by our constant $\frac{1}{n-1}$, we get:

$$I = \frac{1}{n-1} \cdot \frac{n}{n+1} t^{\frac{n+1}{n}} + C = \frac{n}{n^2 - 1} t^{\frac{n+1}{n}} + C$$

Finally, substitute back $t = 1 - \frac{1}{\sin^{n-1} 	heta}$:

$$I = \frac{n}{n^2 - 1} (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$$

And there you have it! We started with a terrifying expression and ended with a clean, elegant result. Remember, in calculus, the path is often hidden, but the tools—factoring, substitution, and patience—will always lead you home.

Animated Solution for Mathematics - Indefinite Integration: Let $n \geq 2$ be a natural number and $0 < θ < π /2$ . Then $\int \frac{( s i n ^{n} θ - s i n θ ) ^{1/ n} c o s θ}{s i n ^{n + 1} θ} d θ$ is equal to : (Where C is a constant of integration)

Select Answer:

Visualized Solution

Analyze the Integrand

Factor out $sin^{n} θ$

Simplify the Inner Fraction

Apply the Power $\frac{1}{n}$

Update the Integral

Simplify the Denominator

Choose the Substitution $t$

Differentiate $t$

Simplify the Derivative

Isolate the Integral Term

Substitute into the Integral

Integrate using Power Rule

Simplify the Coefficient

Back-Substitution

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Power of Factoring

The Simplification Dance

The Substitution

Final Calculation

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