use a change of variables or the table to evaluate the following definite integral.\n int_{1/5}^{sqrt{2}/5}…

use a change of variables or the table to evaluate the following definite integral.\n int_{1/5}^{sqrt{2}/5} \frac{dx}{xsqrt{25x^{2}-1}} \nclick to view the table of general integration formulas.\n int_{1/5}^{sqrt{2}/5} \frac{dx}{xsqrt{25x^{2}-1}}=\text{(type an exact answer.)}

use a change of variables or the table to evaluate the following definite integral.\n int_{1/5}^{sqrt{2}/5} \frac{dx}{xsqrt{25x^{2}-1}} \nclick to view the table of general integration formulas.\n int_{1/5}^{sqrt{2}/5} \frac{dx}{xsqrt{25x^{2}-1}}=\text{(type an exact answer.)}

Answer

Explanation:

Step1: Use substitution

Let $u = 5x$, then $du=5dx$. When $x = \frac{1}{5}$, $u = 1$; when $x=\frac{\sqrt{2}}{5}$, $u=\sqrt{2}$. Also, the integral $\int\frac{dx}{x\sqrt{25x^{2}-1}}$ becomes $\int\frac{du}{u\sqrt{u^{2}-1}}$.

Step2: Recall integral formula

The integral $\int\frac{du}{u\sqrt{u^{2}-1}}=\text{arcsec}(u)+C$.

Step3: Evaluate definite - integral

$\int_{1}^{\sqrt{2}}\frac{du}{u\sqrt{u^{2}-1}}=\left[\text{arcsec}(u)\right]_{1}^{\sqrt{2}}=\text{arcsec}(\sqrt{2})-\text{arcsec}(1)$. Since $\text{arcsec}(\sqrt{2})=\frac{\pi}{4}$ and $\text{arcsec}(1) = 0$, the value of the integral is $\frac{\pi}{4}$.

Answer:

$\frac{\pi}{4}$