int_(0)^(pi//20)20 tan 5xdx

Set Substitution u=5x: We need to find the integral of 20 tan(5x) with respect to x from 0 to π/20. To do this, we will use a substitution method. Let's set u = 5x, which means du/dx = 5 or du = 5dx. We can then solve for dx, which gives us dx = du/5.Substitute u and dx: Now we substitute u and dx in the integral. The integral becomes: int_(0)^(pi//20) 20 tan(5x)dx = int_(0)^(pi//4) 20 tan(u) * (1/5)du This simplifies to: 4 int_(0)^(pi//4) tan(u)duChange Limits of Integration: The limits of integration need to be changed according to our substitution. When x = 0, u = 5*0 = 0. When x = π/20, u = 5*(π/20) = π/4. So the new limits of integration are from 0 to π/4.Evaluate Integral of tan(u): The integral of tan(u) with respect to u is ln|sec(u)|. So we have: 4 int_(0)^(pi//4) tan(u)du = 4[ln|sec(u)|] evaluated from 0 to π/4.Evaluate Antiderivative at Limits: Now we evaluate the antiderivative at the upper and lower limits of integration: 4[ln|sec(π/4)| - ln|sec(0)|] We know that sec(π/4) = √2 and sec(0) = 1, so this becomes: 4[ln(√2) - ln(1)]Simplify Final Expression: Since ln(1) = 0, the expression simplifies to: 4ln(√2) = 4ln(2^(1/2)) = 4*(1/2)ln(2) = 2ln(2)Therefore, the value of the integral from 0 to π/20 of 20 tan(5x) dx is 2ln(2).

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$\int_{0}^{\frac{\pi}{20}} 20 \tan 5x \, dx$

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Evaluate definite integrals using the chain rule

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\int_{0}^{\frac{\pi}{20}} 20 \tan 5x \, dx

Set Substitution $u=5x$ : We need to find the integral of $20 \tan(5x)$ with respect to $x$ from $0$ to $\pi/20$ . To do this, we will use a substitution method. Let's set $u = 5x$ , which means $\frac{du}{dx} = 5$ or $du = 5dx$ . We can then solve for $dx$ , which gives us $dx = \frac{du}{5}$ .
Substitute $u$ and $dx$ : Now we substitute $u$ and $dx$ in the integral. The integral becomes: $\newline$ $\int_{0}^{\pi/20} 20 \tan(5x)\,dx = \int_{0}^{\pi/4} 20 \tan(u) \cdot \left(\frac{1}{5}\right)du$ $\newline$ This simplifies to: $\newline$ $4 \int_{0}^{\pi/4} \tan(u)\,du$
Change Limits of Integration: The limits of integration need to be changed according to our substitution. When $x = 0$ , $u = 5\times 0 = 0$ . When $x = \frac{\pi}{20}$ , $u = 5\times \left(\frac{\pi}{20}\right) = \frac{\pi}{4}$ . So the new limits of integration are from $0$ to $\frac{\pi}{4}$ .
Evaluate Integral of $tan(u)$ : The integral of $tan(u)$ with respect to $u$ is $\ln|\sec(u)|$ . So we have: $4 \int_{0}^{\pi/4} tan(u)\,du = 4[\ln|\sec(u)|]$ evaluated from $0$ to $\pi/4$ .
Evaluate Antiderivative at Limits: Now we evaluate the antiderivative at the upper and lower limits of integration: $\newline$ $4[\ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|]$ $\newline$ We know that $\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\sec(0) = 1$ , so this becomes: $\newline$ $4[\ln(\sqrt{2}) - \ln(1)]$
Simplify Final Expression: Since $\ln(1) = 0$ , the expression simplifies to: $\newline$ $4\ln(\sqrt{2}) = 4\ln(2^{1/2}) = 4\times(1/2)\ln(2) = 2\ln(2)$
Simplify Final Expression: Since $\ln(1) = 0$ , the expression simplifies to: $\newline$ $4\ln(\sqrt{2}) = 4\ln(2^{1/2}) = 4*(1/2)\ln(2) = 2\ln(2)$ Therefore, the value of the integral from $0$ to $\pi/20$ of $20 \tan(5x) \, dx$ is $2\ln(2)$ .