<p>If \(f(x)=\displaystyle\int_0^x t\sin(x-t)\,dt\), find \(f''(x)\). [JEE Main 2017]</p>
Step-by-Step Solution
Key Concept: Use the Laplace convolution trick: f(x) = (t * sin t)(x) = \int_0^x t \cdot sin(x-t)dt = sin x - x cos x. Then differentiate twice.
Step 1: Rewrite the integral using a substitution.
Let $u = x-t$. Then $t = x-u$, and $dt = -du$.
When $t=0$, $u=x$. When $t=x$, $u=0$.
Thus,
$$f(x) = \int_x^0 (x-u) \sin(u) (-du) = \int_0^x (x-u) \sin(u) du$$
$$f(x) = x \int_0^x \sin(u) du - \int_0^x u \sin(u) du$$
Step 2: Evaluate the integrals.
$$f(x) = x [-\cos(u)]_0^x - [-u \cos(u) + \sin(u)]_0^x$$
$$f(x) = x (-\cos(x) - (-\cos(0))) - (-x \cos(x) + \sin(x) - (-0 \cos(0) + \sin(0)))$$
$$f(x) = x (1-\cos(x)) - (-x \cos(x) + \sin(x))$$
$$f(x) = x - x \cos(x) + x \cos(x) - \sin(x)$$
$$f(x) = x - \sin(x)$$
Step 3: Find the first derivative, $f'(x)$.
$$f'(x) = \frac{d}{dx} (x - \sin(x)) = 1 - \cos(x)$$
Step 4: Find the second derivative, $f''(x)$.
$$f''(x) = \frac{d}{dx} (1 - \cos(x)) = 0 - (-\sin(x)) = \sin(x)$$
Correct Answer: A