Question

Let ${\int }_0^x\left(\frac{bt\cos 4t-a\sin 4t}{t^2}\right)dt=\frac{a\sin 4x}{x}$ then $a$ and $b$ are given by

• A. $a=\frac{1}{4},b=1$
• B. $a=2,b=2$
• C. $a=-1,b=4$
• D. $a=2,b=4$

Answer 1

The correct option is A $a=\frac{1}{4},b=1$

${\int }_0^x\frac{bt\cos 4t-a\sin 4t}{t^2}dt=\frac{a\sin 4x}{x}$

$\text{Differentiating w.r.t. x}$

$\frac{bx\cos 4x-a\sin 4x}{x^2}=\frac{4ax\cos 4x-a\sin 4x}{x^2}$

$b=4a\Rightarrow a=\frac{1}{4},b=1$