Question

\(∫sin^2 πx dx =\)?

• A. \(\dfrac{x}{2}-\dfrac{1}{4π} sin2πx+C\)
• B. \(\dfrac{x}{2}+\dfrac{1}{8π} sin4πx+C\)
• C. \(\dfrac{x}{8}-\dfrac{1}{4π} cos2πx+C\)
• D. \(x+\dfrac{1}{2π} sin2πx+C\)
• E. \(\dfrac{x}{2}-\dfrac{1}{2π} cos2πx+C\)

Answer 1

The Correct Option is A

Step 1: Use the power-reduction identity for \(\sin^2 \theta\): \[ \sin^2 \pi x = \frac{1 - \cos(2\pi x)}{2} \]

Step 2: Rewrite the integral: \[ \int \sin^2 \pi x \, dx = \int \frac{1 - \cos(2\pi x)}{2} \, dx \]

Step 3: Separate the integral: \[ \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(2\pi x) \, dx \]

Step 4: Integrate each term: \[ \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin(2\pi x)}{2\pi} + C \]

Step 5: Simplify the expression: \[ \frac{x}{2} - \frac{1}{4\pi} \sin(2\pi x) + C \]

Conclusion: The correct answer is \(\boxed{A}\) \(\left(\frac{x}{2} - \frac{1}{4\pi}\sin(2\pi x) + C\right)\).

Answer 2

We need to evaluate the integral:

\[ \int \sin^2(\pi x) \, dx \]

We'll use the trigonometric identity:

\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]

Applying this identity to our integral:

\[ \int \left[\frac{1 - \cos(2\pi x)}{2}\right] dx = \frac{1}{2} \int \left[1 - \cos(2\pi x)\right] dx \]

Now, integrate term by term:

\[ \frac{1}{2} \left[\int 1 \, dx - \int \cos(2\pi x) \, dx\right] = \frac{1}{2} \left[x - \frac{1}{2\pi} \sin(2\pi x)\right] + C \]

Therefore, the integral evaluates to:

\[ \frac{x}{2} - \frac{1}{4\pi} \sin(2\pi x) + C \]