Evaluate: $\int_{0}^{\frac{π}{4}} log (1 + tan (x)) \cdot d x$ $int_0^(pi/4)log(1+tan(x))*dx$ ?

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Evaluate: $\int_{0}^{\frac{π}{4}} log (1 + tan (x)) \cdot d x$ $int_0^(pi/4)log(1+tan(x))*dx$ ?

The answer is $\frac{π}{2}$ $pi/2$ , but how?

1 Answer

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Answer 1 · 2017-11-14T05:31:11

$\int_{0}^{\frac{π}{4}} log (1 + tan x) d x = \frac{π}{8} log 2$ $int_0^(pi/4)log(1+tanx)dx=pi/8log2$

Explanation:

As $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ $int_0^af(x)dx=int_0^af(a-x)dx$

Let $I = \int_{0}^{\frac{π}{4}} log (1 + tan x) d x = \int_{0}^{\frac{π}{4}} log (1 + tan (\frac{π}{4} - x)) d x$ $I=int_0^(pi/4)log(1+tanx)dx=int_0^(pi/4)log(1+tan(pi/4-x))dx$

= $\int_{0}^{\frac{π}{4}} log (1 + \frac{1 - tan x}{1 + tan x}) d x$ $int_0^(pi/4)log(1+(1-tanx)/(1+tanx))dx$

= $\int_{0}^{\frac{π}{4}} log (\frac{2}{1 + tan x}) d x$ $int_0^(pi/4)log(2/(1+tanx))dx$

= $\int_{0}^{\frac{π}{4}} log 2 d x - \int_{0}^{\frac{π}{4}} log (1 + tan x) d x$ $int_0^(pi/4)log2dx-int_0^(pi/4)log(1+tanx)dx$

= $\int_{0}^{\frac{π}{4}} log 2 d x - I$ $int_0^(pi/4)log2dx-I$

Hence $2 I = \int_{0}^{\frac{π}{4}} log 2 d x$ $2I=int_0^(pi/4)log2dx$

and $I = log 2 \times \frac{π}{4} \times \frac{1}{2} = \frac{π}{8} log 2$ $I=log2xxpi/4xx1/2=pi/8log2$

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Evaluate: $\int_{0}^{\frac{π}{4}} log (1 + tan (x)) \cdot d x$ $int_0^(pi/4)log(1+tan(x))*dx$ ?

The answer is $\frac{π}{2}$ $pi/2$ , but how?

1 Answer

Explanation:

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Evaluate: $\int_{0}^{\frac{π}{4}} log (1 + tan (x)) \cdot d x$ $int_0^(pi/4)log(1+tan(x))*dx$ ?

The answer is $\frac{π}{2}$ $pi/2$ , but how?