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∫ln(tanx) dx limit is from 0 to pi/4

∫ln(tanx) dx
limit is from 0 to pi/4

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 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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