∫1/331xsin1x−xdx=.....

Put 1x−x=t. Then −1x2−1dx=dt⇒⋅dxx1x+x=dt⇒−dxx1x−x2+4=dt⇒−dxx=dtt2+4⇒dxx=−dtt2+4 and x=13,3⇒t=83,−83∴∫1/311xsin1x−xdx=∫8/3−8/3sint−dtt2+4=∫8/3−8/3sintt2+4dt=0.

∫1/331xsin1x−xdx=.....

Put 1x−x=t. Then −1x2−1dx=dt⇒⋅dxx1x+x=dt⇒−dxx1x−x2+4=dt⇒−dxx=dtt2+4⇒dxx=−dtt2+4 and x=13,3⇒t=83,−83∴∫1/311xsin1x−xdx=∫8/3−8/3sint−dtt2+4=∫8/3−8/3sintt2+4dt=0.

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$\int_{1 / 3}^{3} \frac{1}{x} \sin (\frac{1}{x} - x) d x = .....$

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Detailed Solution

Put $\frac{1}{x} - x = t . T h e n (- \frac{1}{x^{2}} - 1) d x = d t \Rightarrow \cdot \frac{d x}{x} (\frac{1}{x} + x) = d t \Rightarrow - \frac{d x}{x} \sqrt{{(\frac{1}{x} - x)}^{2} + 4} = d t$

$\Rightarrow - \frac{d x}{x} = \frac{d t}{\sqrt{t^{2} + 4}} \Rightarrow \frac{d x}{x} = - \frac{d t}{\sqrt{t^{2} + 4}} a n d x = \frac{1}{3}, 3 \Rightarrow t = \frac{8}{3}, - \frac{8}{3}$

$∴ \int_{1 / 3}^{1} \frac{1}{x} \sin (\frac{1}{x} - x) d x = \int_{8 / 3}^{- 8 / 3} \sin t (\frac{- d t}{\sqrt{t^{2} + 4}} =) \int_{8 / 3}^{- 8 / 3} \frac{\sin t}{\sqrt{t^{2} + 4}} d t = 0.$