Question

$\int\frac{x\cos x-\sin x}{x^{2}}dx$ is equal to

• A. $\frac{\cos x}{x}+C$
• B. $\frac{\sin 2x}{x}+C$
• C. $\frac{\sin x}{x^{2}}+C$
• D. $\frac{\sin x}{x}+C$
• E. $\frac{\cos 2x}{x}+C$

Hint:

Logic Tip: Whenever you see an integral with $x^2$ in the denominator and a subtraction in the numerator involving $x$ and trigonometric/logarithmic functions, immediately suspect the reverse Quotient Rule. Look at the term *without* the $x$ multiplier in the numerator to identify the original function $u$.

Answer 1

The Correct Option is D

Concept:
The integrand has a very specific structure that perfectly resembles the result of applying the Quotient Rule for differentiation. The Quotient Rule is given by: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}$$
Step 1: Analyze the structure of the integrand.
The given integrand is $\frac{x\cos x - \sin x}{x^2}$. Comparing this to the quotient rule formula: The denominator is $v^2 = x^2$, which implies $v = x$. The numerator is $v \cdot u' - u \cdot v' = x \cdot u' - u \cdot (1)$. From the given numerator $x\cos x - \sin x$, we can deduce: $u' = \cos x$ $u = \sin x$
Step 2: Verify the deduction.
Let's check if $u = \sin x$ matches $u' = \cos x$: $$\frac{d}{dx}(\sin x) = \cos x$$ This matches perfectly. Therefore, the function that was differentiated to produce this integrand is $\frac{u}{v} = \frac{\sin x}{x}$.
Step 3: Perform the integration.
Since integration is the reverse of differentiation: $$\int \frac{d}{dx} \left( \frac{\sin x}{x} \right) dx = \frac{\sin x}{x} + C$$