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$\int_{0}^{\alpha} \ln(\cot(\alpha) + \tan(x)) \, dx$ , where $\alpha$ is in the interval $0, \frac{\pi}{2}$

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Determine end behavior of polynomial and rational functions

Full solution

Q.

\int_{0}^{\alpha} \ln(\cot(\alpha) + \tan(x)) \, dx

, where

\alpha

is in the interval

0, \frac{\pi}{2}

Simplify integrand: First, let's simplify the integrand $\ln(\cot(\alpha) + \tan(x))$ . $\cot(\alpha)$ is constant with respect to $x$ , so we can take it out of the integral. We get $\int_{0}^{\alpha} \ln(\cot(\alpha)) \, dx + \int_{0}^{\alpha} \ln(\tan(x)) \, dx$ .
Integrate $\ln(\cot(\alpha))$ : Now, let's integrate the first part $\int_{0}^{\alpha} \ln(\cot(\alpha)) \, dx$ . Since $\ln(\cot(\alpha))$ is a constant, the integral is $\ln(\cot(\alpha)) \cdot x$ evaluated from $0$ to $\alpha$ . This gives us $\alpha \cdot \ln(\cot(\alpha))$ .
Integrate $\ln(\tan(x))$ : Next, we integrate the second part $\int_{0}^{\alpha} \ln(\tan(x)) \, dx$ . This is a bit trickier, we can use the substitution method. Let $u = \tan(x)$ , then $du = \sec^2(x) \, dx$ . But we don't have $\sec^2(x)$ in our integral, so this approach doesn't work.

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