Question

The integral $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \text{ is equal to} $

• A. 0
• B. \( \pi \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{4} \)

Hint:

For integrals involving \( \sin^2 x \), you can use the identity \( \sin^2 x = \frac{1}{2} (1 - \cos(2x)) \) to simplify the expression before integrating.

Answer 1

The Correct Option is C

To solve the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \), we can use the trigonometric identity: \(\sin^2 x = \frac{1-\cos(2x)}{2}\).

This allows us to rewrite the integral as:

\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1-\cos(2x)}{2} \, dx\]

Breaking it down, we have:

\[\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx\]

Calculating the first part:

\[\frac{1}{2} \left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx\right] = \frac{1}{2} \left[x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2}(\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi}{2}\]

Calculating the second part:

\[\frac{1}{2} \left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx\right]\]

The integral of \(\cos(2x)\) is \(\frac{1}{2}\sin(2x)\). Therefore:

\[\frac{1}{2}\left[\frac{1}{2}\sin(2x) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{4}\left[\sin(\pi) - \sin(-\pi)\right] = \frac{1}{4}(0 - 0) = 0\]

Putting it all together:

\[\frac{\pi}{2} - 0 = \frac{\pi}{2}\]

Thus, the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{2}\).

Answer 2

To evaluate the integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \] We can use the identity \( \sin^2 x = \frac{1}{2} (1 - \cos(2x)) \) to simplify the integrand: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (1 - \cos(2x)) \, dx \] This separates into two integrals: \[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \] The first integral evaluates to: \[ \frac{1}{2} \left( x \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2} \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) = \frac{\pi}{2} \] The second integral is: \[ \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \] The integral of \( \cos(2x) \) is \( \frac{1}{2} \sin(2x) \), and since \( \sin(2x) \) is zero at both bounds \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), this part evaluates to zero: \[ \frac{1}{2} \left[ \frac{1}{2} \sin(2x) \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right] = 0 \] Thus, the total value of the integral is: \[ I = \frac{\pi}{2} \]