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Evaluate.
(k lambda)/(4R)int_(-(pi)/(2))^(pi//2)sec ((theta)/(2))*d theta

Evaluate.\newlinekλ4Rπ2π/2secθ2dθ\frac{k \lambda}{4 R} \int_{-\frac{\pi}{2}}^{\pi / 2} \sec \frac{\theta}{2} \cdot d \theta

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Q. Evaluate.\newlinekλ4Rπ2π/2secθ2dθ\frac{k \lambda}{4 R} \int_{-\frac{\pi}{2}}^{\pi / 2} \sec \frac{\theta}{2} \cdot d \theta
  1. Identify Integral: Identify the integral that needs to be evaluated.\newlineWe have the integral (π2)π2sec(θ2)dθ\int_{-\left(\frac{\pi}{2}\right)}^{\frac{\pi}{2}} \sec\left(\frac{\theta}{2}\right) d\theta with limits from (π2)-\left(\frac{\pi}{2}\right) to π2\frac{\pi}{2}.
  2. Simplify Integrands: Use a trigonometric identity to simplify the integrand. The secant function can be expressed as sec(x)=1cos(x)\sec(x) = \frac{1}{\cos(x)}. Therefore, sec(θ2)=1cos(θ2)\sec(\frac{\theta}{2}) = \frac{1}{\cos(\frac{\theta}{2})}.
  3. Recognize Integral Rule: Recognize that the integral of sec(x)\sec(x) is lnsec(x)+tan(x)+C\ln|\sec(x) + \tan(x)| + C. However, since we have sec(θ/2)\sec(\theta/2), we need to adjust the integral accordingly.
  4. Perform Substitution: Perform a substitution to integrate sec(θ2)\sec(\frac{\theta}{2}). Let u=θ2u = \frac{\theta}{2}, which implies dθ=2dud\theta = 2du. We need to change the limits of integration as well. When θ=(π2)\theta = -\left(\frac{\pi}{2}\right), u=(π4)u = -\left(\frac{\pi}{4}\right), and when θ=π2\theta = \frac{\pi}{2}, u=π4u = \frac{\pi}{4}.
  5. Rewrite in Terms of u: Rewrite the integral in terms of u.\newlineThe integral becomes π4π4sec(u)2du\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec(u) \cdot 2\,du with limits from π4-\frac{\pi}{4} to π4\frac{\pi}{4}.
  6. Evaluate with New Limits: Evaluate the integral with the new limits.\newlineπ4π4sec(u)2du=2lnsec(u)+tan(u)\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec(u) \cdot 2\,du = 2 \cdot \ln|\sec(u) + \tan(u)| from π4-\frac{\pi}{4} to π4\frac{\pi}{4}.
  7. Calculate Definite Integral: Calculate the definite integral. \newline2×[lnsec(π4)+tan(π4)lnsec(π4)+tan(π4)]2 \times [\ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(-\frac{\pi}{4}) + \tan(-\frac{\pi}{4})|].
  8. Simplify Using Trig Functions: Simplify the expression using the fact that sec(π4)=2\sec(\frac{\pi}{4}) = \sqrt{2} and tan(π4)=1\tan(\frac{\pi}{4}) = 1 (and similarly for π4-\frac{\pi}{4}).\newline2[ln(2+1)ln(21)]2 \cdot [\ln(\sqrt{2} + 1) - \ln(\sqrt{2} - 1)].
  9. Combine Logarithmic Terms: Use the properties of logarithms to combine the terms. 2×ln[(2+1)(21)].2 \times \ln\left[\frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1)}\right].
  10. Rationalize Denominator: Rationalize the denominator of the argument of the logarithm.\newline2ln((2+1)(2+1)(21)(2+1))=2ln(2+22+121)=2ln(3+22)2 \cdot \ln\left(\frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)}\right) = 2 \cdot \ln\left(\frac{2 + 2\sqrt{2} + 1}{2 - 1}\right) = 2 \cdot \ln(3 + 2\sqrt{2}).
  11. Multiply by Constant Factor: Multiply the result by the constant factor (kλ4R)(\frac{k \lambda}{4R}).kλ4R×2×ln(3+22)=kλ2R×ln(3+22)\frac{k \lambda}{4R} \times 2 \times \ln(3 + 2\sqrt{2}) = \frac{k \lambda}{2R} \times \ln(3 + 2\sqrt{2}).

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