Question

The indefinite integral of \( \sin(x) \) w.r.t \( \cos(x) \) is:

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

Hint:

Using substitution \( u = \cos(x) \) makes it very simple:
\[ \int \sin(x) \, du = \int \sqrt{1 - u^2} \, du \] Alternatively, converting \( d(\cos(x)) \) directly to \( -\sin(x)dx \) is often faster and less prone to algebraic errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem requires evaluating the indefinite integral of the function \( \sin(x) \) with respect to \( \cos(x) \).

Step 2: Key Formula or Approach:
The integral of a function \( f(x) \) with respect to another function \( g(x) \) is formulated as:
\[ \int f(x) \, d(g(x)) = \int f(x) \cdot g'(x) \, dx \]

Step 3: Detailed Explanation:
Let \( g(x) = \cos(x) \). Then its differential is:
\[ d(\cos(x)) = -\sin(x) \, dx \] Substitute this into the required integral:
\[ I = \int \sin(x) \, d(\cos(x)) = \int \sin(x) \left(-\sin(x) \, dx\right) \] \[ I = -\int \sin^2(x) \, dx \] Using the trigonometric identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
\[ I = -\int \frac{1 - \cos(2x)}{2} \, dx \] \[ I = -\frac{1}{2} \int (1 - \cos(2x)) \, dx \] \[ I = -\frac{1}{2} \left( x - \frac{\sin(2x)}{2} \right) + c \] \[ I = -\frac{x}{2} + \frac{\sin(2x)}{4} + c \] \[ I = \frac{\sin(2x)}{4} - \frac{x}{2} + c \] This matches option (B).

Step 4: Final Answer:
(B) \( \frac{\sin(2x)}{4} - \frac{x}{2} + c \)