Question

The value of ${\int }_{1/3}^3\frac{\tan (x^2+\frac{1}{x^2})}{\sin (x+\frac{1}{x})}\frac{dx}{x}$ is

Answer 1

Let $I={\int }_{1/3}^3\frac{\tan (x^2+\frac{1}{x^2})}{\sin (x+\frac{1}{x})}\frac{dx}{x}$

Let $x+\frac{1}{x}=t$............. I

When $x=\frac{1}{3},t=\frac{10}{3}$

When $x=3,t=\frac{10}{3}$

Squaring both side, we get

$\Rightarrow x^2+\frac{1}{x^2}=t^2-2$.............. II

Differentiating both sides of equation I, we get

$\Rightarrow (1-\frac{1}{x^2})dx=dt$

$\Rightarrow (x-\frac{1}{x})\frac{dx}{x}=dt$................. III

Now, ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$

$=t^2-2-2$

$=t^2-4$

$\Rightarrow (x-\frac{1}{x})=\sqrt{t^2-4}$................. IV

From equation III and IV, we get

$\Rightarrow \frac{dx}{x}=\frac{dt}{\sqrt{t^2-4}}$.............. V

Substituting equations I, II and V in the equation given in question and changing the limits accordingly, we get

$I={\int }_{10/3}^{10/3}\frac{\tan (t^2-2)dt}{\sqrt{t^2-4}\sin t}$

$=0$ [$\because$ upper and lower limits of integration become equal]

Hence, the answer is $0$