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CBSE Class 12 · 2006 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{π /4} lo g (1 + tan x) d x

Answer: $\frac{π}{8} lo g 2$

Solution

Let

I = \int_{0}^{π /4} lo g (1 + tan x) d x

. Apply

x \to \frac{π}{4} - x

:

1 + tan (\frac{π}{4} - x) = 1 + \frac{1 - tan x}{1 + tan x} = \frac{2}{1 + tan x}

I = \int_{0}^{π /4} lo g \frac{2}{1 + tan x} d x = \int_{0}^{π /4} lo g 2 d x - I

2 I = lo g 2 \cdot \frac{π}{4} \Rightarrow I = \frac{π}{8} lo g 2

CBSE Class 12 Board · 2006 · Subjective

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