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Integral Calculus> ∫ln(tanx) dx limit is from 0 to pi/4...

question mark

∫ln(tanx) dx
limit is from 0 to pi/4

Lovey , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$I = \int_{0}^{\pi /4}ln(1+tanx)dx$
$I = \int_{0}^{\pi /4}ln(1+tan(\frac{\pi }{4}-x))dx$
$I = \int_{0}^{\pi /4}ln(1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx})dx$
$I = \int_{0}^{\pi /4}ln(1+\frac{1-tanx}{1+tanx})dx$
$I = \int_{0}^{\pi /4}ln(\frac{2}{1+tanx})dx$
$I = \int_{0}^{\pi /4}(ln(2)-ln(1+tanx))dx$
$I = \int_{0}^{\pi /4}ln(2)dx-\int_{0}^{\pi /4}ln(1+tanx)dx$
$I = \int_{0}^{\pi /4}ln(2)dx-I$
$2I = \int_{0}^{\pi /4}ln(2)dx$
$2I = ln(2)[x]_{0}^{\pi /4}$
$I = \frac{\pi }{8}ln(2)$

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