<s>[INST] $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac{3}{2}$ but what to do with remaining $\frac{1}{x}$ and $dx$ in the integration.Please help....
 [/INST] The fact that the limits are reciprocals of each other gives you your first clue. Splitting the integral up about the geometric mean $\sqrt{2\cdot\frac 12} = 1$, we have
$$I = \int_{1/2}^1 \frac{1}{x}\sin (x-1/x) \ dx + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx$$
Now change variables in the first of those two integrals by $t = 1/x$ say, and we have
$$\int_2^1 \frac{t}{-t^2} \sin(1/t - t) \ dt = \int_2^1 \frac{1}{t} \sin(t - 1/t) \ dt = - \int_1^2 \frac{1}{t} \sin(t - 1/t) \ dt$$
Thus
$$I = - \int_1^2 \frac{1}{t} \sin(t - 1/t) \ dt + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx \ = \ \cdots$$
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