Nann

2021-10-10

Evaluate the integrals.

$\int}_{\sqrt{2}}^{2}\frac{dx}{x\sqrt{{x}^{2}-1}$

Brighton

Skilled2021-10-11Added 103 answers

Evaluate,

$\int}_{\sqrt{2}}^{2}\frac{dx}{x\sqrt{{x}^{2}-1}$

Simplification:

Since we have,

$\int}_{\sqrt{2}}^{2}\frac{dx}{x\sqrt{{x}^{2}-1}$

Substitute

$\sqrt{{x}^{2}-1}=u\Rightarrow \frac{2x}{2\sqrt{{x}^{2}-1}}dx=du\Rightarrow \frac{1}{\sqrt{{x}^{2}-1}}dx=\frac{1}{x}du$

When$x=\sqrt{2},u=1$

When$x=2,u=3$

In (1),

$\int}_{\sqrt{2}}^{2}\frac{dx}{x\sqrt{{x}^{2}-1}}={\int}_{1}^{\sqrt{3}}\frac{du}{{x}^{2}$

$={\int}_{1}^{\sqrt{3}}\frac{du}{{x}^{2}}$

$={\int}_{1}^{\sqrt{3}}\frac{du}{1+{u}^{2}}$

$={\left[{\mathrm{tan}}^{-1}u\right]}_{1}^{\sqrt{3}}$

$=[{\mathrm{tan}}^{-1}\sqrt{3}-{\mathrm{tan}}^{-1}1]$

$=[\frac{\pi}{3}-\frac{\pi}{4}]$

$=\left[\frac{4\pi -3\pi}{12}\right]$

$=\frac{\pi}{12}$

Hence

$\int}_{\sqrt{2}}^{2}\frac{dx}{x\sqrt{{x}^{2}-1}}=\frac{\pi}{12$

Simplification:

Since we have,

Substitute

When

When

In (1),

Hence

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