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: Translate the unusual notation.
Integrating $\sin x$ with respect to $\cos x$ means $\displaystyle\int \sin x\,d(\cos x)$.
Step 2: Substitute the differential.
Since $d(\cos x) = -\sin x\,dx$, the integral becomes $\displaystyle\int \sin x\,(-\sin x)\,dx = -\int \sin^2 x\,dx$.
Step 3: Apply the substitution $t=\cos x$ to double-check.
With $t=\cos x$, $\sin^2 x = 1-t^2$, so $-\int \sin^2x\,dx = -\int (1-t^2)\,\dfrac{dt}{-\sin x}$; this loops back, so we instead use the power-reduction identity directly.
Step 4: Power-reduce the integrand.
Using $\sin^2 x = \dfrac{1-\cos 2x}{2}$, \[ -\int \sin^2 x\,dx = -\frac{1}{2}\int (1-\cos 2x)\,dx \]
Step 5: Integrate term by term.
\[ = -\frac{1}{2}\left(x - \frac{\sin 2x}{2}\right) + c = -\frac{x}{2} + \frac{\sin 2x}{4} + c \]
Step 6: Tidy the order.
Rearranging gives the standard form. \[ \boxed{\dfrac{\sin 2x}{4} - \dfrac{x}{2} + c} \]