Integrate log(1 + tan x) from 0 to π/4.
We will be using a definite integral property for solving this question.
Answer: Integration of log(1 + tan x) from 0 to π/4 is equal to π/8 log 2.
Let's integrate.
Explanation:
Let I = \(\int_{0}^{\pi/4}\)log(1 + tan x) dx
Since, this is a definite integral, to integrate it we have to use the following property of definite integrals.
\(\int_{0}^{a}\) f(x)dx = \(\int_{0}^{a}\)f(a - x) dx
Therefore,
I = \(\int_{0}^{\pi / 4} \) log (1 + tan[(π /4)-x] dx ---- (1)
Using, tan(a - b) = (
I = \(\int_{0}^{\pi / 4}\) log[1 + (tan(π/4) - tan x) / (1 + tan(π/4) tan x) ]dx
I = \(\int_{0}^{\pi / 4}\) log[1+ (1 - tan x)/(1 + tan x) ]dx
I = \(\int_{0}^{\pi / 4}\) log[1 + (1 + tan x - tan x) / (1 + tan x)]dx
I = \(\int_{0}^{\pi / 4}\) log[2/(1+ tan x) ]dx
Using log (a/b) = log a – log b
I = \(\int_{0}^{\pi / 4}\) (log 2 - log (1 + tan x))dx
I = \(\int_{0}^{\pi / 4}\) log 2 dx - \(\int_{0}^{\pi / 4}\)
Adding equation (1) and equation (2), we get
I + I = \(\int_{0}^{\pi / 4} \) log (1 + tan[(π /4) - x]) dx + \(\int_{0}^{\pi / 4}\) log 2 dx - \(\int_{0}^{\pi / 4}\) log (1 + tan x) dx
2I = \(\int_{0}^{\pi / 4} \) log 2 dx
2I = log 2 \(\int_{0}^{\pi / 4} \)dx
I = (log 2)/2
I = (log 2)/2 (π/4 - 0)
I = (log 2)/2 × (π/4)
I = (π/8) log 2
Use Online Cuemath's Definite Integral Calculator to find the definite integration.
Hence, the integral of log(1 + tan x) from 0 to π/4 is (π/8) log 2.
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