Question

\(\int \sin^2 x \,dx =\)

• A. \(\frac{x}{2} + \frac{\sin 2x}{4} + c\)
• B. \(\frac{x}{2} - \frac{\cos 2x}{4} + c\)
• C. \(\frac{x}{2} + \frac{\cos 2x}{4} + c\)
• D. \(\frac{x}{2} - \frac{\sin 2x}{4} + c\)

Hint:

Memorize the power-reduction formulas for \(\sin^2 x\) and \(\cos^2 x\), as they are fundamental for integrating even powers of sine and cosine.

• \(\sin^2 x = \frac{1 - \cos(2x)}{2}\)

• \(\cos^2 x = \frac{1 + \cos(2x)}{2}\)
These are direct applications of the \(\cos(2x)\) double-angle identities.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the indefinite integral of \(\sin^2 x\).

Step 2: Key Formula or Approach:
We cannot integrate \(\sin^2 x\) directly. We must first reduce the power using a trigonometric identity. The relevant power-reduction formula comes from the double-angle identity for cosine:
\[ \cos(2x) = 1 - 2\sin^2 x \]
Rearranging this formula to solve for \(\sin^2 x\), we get:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]

Step 3: Detailed Explanation:
Let \(I = \int \sin^2 x \,dx\).
Substitute the power-reduction formula into the integral:
\[ I = \int \frac{1 - \cos(2x)}{2} \,dx \]
We can split this into two separate integrals:
\[ I = \frac{1}{2} \int (1 - \cos(2x)) \,dx = \frac{1}{2} \left( \int 1 \,dx - \int \cos(2x) \,dx \right) \]
Integrate each term:
\[ \int 1 \,dx = x \]
\[ \int \cos(2x) \,dx = \frac{\sin(2x)}{2} \]
Now, substitute these back into the expression for \(I\):
\[ I = \frac{1}{2} \left( x - \frac{\sin(2x)}{2} \right) + c \]
Distribute the \(\frac{1}{2}\):
\[ I = \frac{x}{2} - \frac{\sin(2x)}{4} + c \]

Step 4: Final Answer:
The integral of \(\sin^2 x\) is \(\frac{x}{2} - \frac{\sin 2x}{4} + c\).