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# 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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