Let F(x) be an indefinite integral of sin²x. STATEMENT-1: The function F(x) satisfies F(x + π) = F(x

Q: If $\sin^2 x$ is periodic, why isn't its integral periodic?

This is a classic trap! Periodicity of a function does not guarantee the periodicity of its integral. The integral represents the accumulation of area. If the average value of the function over one period is non-zero, the integral will accumulate a net area, causing it to drift linearly (like the $\frac{x}{2}$ term here) rather than repeating.

Q: Can I just use the power rule for $\int \sin^2 x \, dx$?

No, the power rule $\int x^n dx$ only applies to polynomials. For trigonometric powers, you must use identities like the double-angle formula to linearize the expression before integrating.

Q: Does the constant of integration $C$ affect the periodicity?

Great question! No, $C$ is just a vertical shift. It changes the graph's position on the y-axis but does not change its shape or its periodic nature. The lack of periodicity here is caused by the linear $x$ term, not the constant $C$.

Question

Let $F(x)$ be an indefinite integral of $\sin^2 x$. \ STATEMENT-1: The function $F(x)$ satisfies $F(x + \pi) = F(x)$ because \ STATEMENT-2: $\sin^2(x + \pi) = \sin^2 x$ for all real $x$.

Options:

(A) Statement-1 is True, statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical landscape. Today, we are diving into a problem that feels simple on the surface but hides a profound conceptual lesson.

We are investigating the relationship between a periodic function and its indefinite integral. It is a classic JEE Advanced scenario: two statements, one logical trap, and a beautiful mathematical resolution. Let us begin by examining the integrand, $f(x) = \sin^2 x$.

### Phase 1: The Integrand

Statement-2 posits that $\sin^2 x$ is periodic with a period of $\pi$. To verify this, we test the definition of periodicity: $f(x + \pi) = f(x)$.

Substituting $x + \pi$ into our function, we get $\sin^2(x + \pi)$. Recalling the ASTC rule from your trigonometry toolkit, we know that $\sin(x + \pi) = -\sin x$.

Squaring this gives $(-\sin x)^2 = \sin^2 x$. Thus, $f(x + \pi) = f(x)$. Statement-2 is undeniably true; the function $\sin^2 x$ repeats its values every $\pi$ units.

### Phase 2: The Integral

Now, we turn to Statement-1, which claims that the indefinite integral $F(x) = \int \sin^2 x \, dx$ is also periodic with period $\pi$. This is where many students stumble.

They assume that if a function is periodic, its integral must be too. But let us perform the integration to see the truth. We must use the double-angle identity: $\cos 2x = 1 - 2\sin^2 x$, which rearranges to:

$$\sin^2 x = \frac{1 - \cos 2x}{2}$$

Substituting this into our integral, we get:

$$F(x) = \int \frac{1 - \cos 2x}{2} \, dx$$

Splitting this into two parts, we have:

$$F(x) = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos 2x \, dx$$

Performing the integration, we arrive at:

$$F(x) = \frac{x}{2} - \frac{\sin 2x}{4} + C$$

### Phase 3: The Comparison

Now, let us test the periodicity of $F(x)$ by substituting $x + \pi$ into our result:

$$F(x + \pi) = \frac{x + \pi}{2} - \frac{\sin 2(x + \pi)}{4} + C$$

Expanding the arguments, we get:

$$F(x + \pi) = \frac{x}{2} + \frac{\pi}{2} - \frac{\sin(2x + 2\pi)}{4} + C$$

Since $\sin(2x + 2\pi) = \sin 2x$, this simplifies to:

$$F(x + \pi) = \frac{x}{2} + \frac{\pi}{2} - \frac{\sin 2x}{4} + C$$

Rearranging to isolate our original $F(x)$, we find:

$$F(x + \pi) = \left( \frac{x}{2} - \frac{\sin 2x}{4} + C ight) + \frac{\pi}{2}$$

This results in the final relation:

$$F(x + \pi) = F(x) + \frac{\pi}{2}$$

### The Verdict

The result is clear: $F(x + \pi) 
eq F(x)$. The presence of the linear term $\frac{x}{2}$ means that for every period $\pi$, the function shifts upwards by $\frac{\pi}{2}$.

It does not repeat; it drifts. Therefore, **Statement-1 is false**.

This problem teaches us that while the *rate of change* (the derivative) of a periodic function is periodic, the *accumulation* (the integral) is only periodic if the average value over a period is zero. Keep this distinction in your heart, and you will never fall for this trap again.

Animated Solution for Mathematics - Indefinite Integration: Let $F (x)$ be an indefinite integral of $sin^{2} x$ . \\ STATEMENT-1: The function $F (x)$ satisfies $F (x + π) = F (x)$ because \\ STATEMENT-2: $sin^{2} (x + π) = sin^{2} x$ for all real $x$ .

Select Answer:

Visualized Solution

Analyzing the Statements

Evaluating Statement-2

Trigonometric Substitution

Conclusion for Statement-2

Integrating $f (x)$

Using the Half-Angle Identity

Setting up the Integral

Performing the Integration

Testing Periodicity of $F (x)$

Substituting $x + π$

Simplifying the Trigonometric Term

Comparing $F (x + π)$ and $F (x)$

The Final Verdict

Conclusion

The Sigma Insight: Fundamental Integration Formulas

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Verdict

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