$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ 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....
 A: Start with $\dfrac1x=t$ to get $$I=\int_2^{1/2}t\sin\left(\dfrac1t-t\right)\left(\dfrac{dt}{-t^2}\right)$$
$$=-\int_2^{1/2}\dfrac1t\sin\left(\dfrac1t-t\right)dt$$
$$=\int_{1/2}^2\dfrac1t\sin\left(\dfrac1t-t\right)dt\text{ (as }\displaystyle\int_a^bf(x)\ dx=-\int_b^af(x)\ dx )$$
$$=-\int_{1/2}^2\dfrac1t\sin\left(t-\dfrac1t\right)dt\text{ (as }\displaystyle \sin(-u)=-\sin u)$$
$$\implies I=-I$$
A: 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
$$I = - \int_1^2 \frac{1}{t} \sin(t - 1/t) \ dt + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx \ = \ \cdots$$
A: $\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \sin \left(x-\frac{1}{x}\right)dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt\Rightarrow \left(x+\frac{1}{x}\right)dx = xdt$ 
and Changing Limits
Now Using $$\displaystyle \left(x+\frac{1}{x}\right)^2 -\left(x-\frac{1}{x}\right)^2=4\Rightarrow \left(x-\frac{1}{x}\right)=\sqrt{\left(x-\frac{1}{x}\right)^2+4}=\sqrt{t^2+4}$$
So Integral $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}\cdot \frac{x}{x}dt = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}dt$$
So we get $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\underbrace{\frac{\sin t}{\sqrt{t^2+4}}}_{\bf{odd\; function}}dt = 0$$
Above we have used the formula $$\displaystyle \int_{-a}^{a}f(x)dx = 0\;,$$ If $f(x)$ is odd function.
A: $\bf{Another\; Solution::}$ Let $$\displaystyle I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \sin \left(x-\frac{1}{x}\right)dx\;,$$ Now Let $x=e^{t}\;,$ Then $dx = e^{t}dt$
and Changing Limits ,
$x=e^{t}\Rightarrow t=\ln(x)\;,$ So $\displaystyle t_{\bf{lower}} = \ln\left(\frac{1}{2}\right)=-\ln(2)$ and $\displaystyle t_{\bf{upper}} = \ln\left(2\right) = \ln(2)$
We get $$\displaystyle I = \int_{-\ln(2)}^{\ln(2)}\frac{1}{e^{t}}\cdot \sin \left(e^{t}-e^{-t}\right)\cdot e^{tdt} = \int_{-\ln(2)}^{\ln(2)}\sin \left(e^{t}-e^{-t}\right)dt$$
So We get $$\displaystyle I = \int_{-\ln(2)}^{\ln(2)}\underbrace{\sin \left(e^{t}-e^{-t}\right)}_{\bf{Odd\; function}}dt = 0$$
A: Let $$I(a) = \int^{a}_{\frac{1}{a}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$$
Where $a>0$
Now $$I'(a) = \frac{1}{a}\sin \left(a-\frac{1}{a}\right)+\sin \left(\frac{1}{a}-a\right)\cdot a \cdot \frac{1}{a^2} = 0$$
So $$I'(a) = 0\Rightarrow I(a) = \mathcal{C}$$
Now put $a=1\;,$ in first equation, We get $I(1) = 0$
So  put  $a = 1$ in $I(a)= \mathcal{C}\;,$ We get $I(1) = 0 = \mathcal{C}$
So we get $\displaystyle I(a) = 0\;,$ Put $a=2\;,$ We get $$I(2) = \int^{2}_{\frac{1}{2}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx = 0$$
