First slide

Evaluation of definite integrals

Question

If the equality
$\int_{0}^{x} \frac{b t \cos 4 t - a \sin 4 t}{t^{2}} d t = \frac{a \sin 4 x}{x} - 1$
holds for all x such that $0 < x < π / 4$ then a and b are given by

Moderate

A

$a = 1 / 4, b = 1$
B

$a = 2, b = 2$
C

$a = - 1, b = 4$
D

$a = 2, b = 4$

Solution

$\int_{0}^{x} \frac{b t \cos 4 t - a \sin 4 t}{t^{2}} d t$
$\begin{array}{l} = b \int_{0}^{x} \frac{\cos (4 t)}{t} d t - a \int_{0}^{x} \frac{\sin (4 t)}{t^{2}} d t = b \int_{0}^{x} \frac{\cos 4 t}{t} d t - a [- {\frac{\sin 4 t}{t}|}_{0}^{x} + 4 \int_{0}^{x} \frac{\cos 4 t}{t} d t] \\ = (b - 4 a) \int_{0}^{x} \frac{\cos 4 t}{t} d t + \frac{a \sin 4 x}{x} - 4 a \end{array}$
Thus $(b - 4 a) \int_{0}^{x} \frac{x \cos 4 t}{t} d t + \frac{a \sin 4 x}{x} - 4 a$
$= \frac{a \sin 4 x}{x} - 1$
i.e. $(b - 4 a) \int_{0}^{x} \frac{x \cos 4 t}{t} d t = 4 a - 1$ .
Since R.H.S. is independent of x, so we must have $b - 4 a = 0$ and $4 a - 1 = 0$ i.e., $a = 1 / 4, b = 1$ .

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