Question

integrate log(1+tanx)​

Answer 1

Answer:

Step-by-step explanation:

Jitender Singh

IIT Delhi

askIITians Faculty

158 Points

Ans:

I = \int_{0}^{\pi /4}ln(1+tanx)dx

I = \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)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)dx-\int_{0}^{\pi /4}ln(1+tan(x))dx

I = \int_{0}^{\pi /4}ln(2)dx-I

2I = \int_{0}^{\pi /4}ln(2)dx

I = \frac{ln(2)}{2}(x)_{0}^{\pi /4}

I = \frac{\pi }{8}ln(2)

Thanks & Regards

Jitender Singh

IIT Delhi

askIITians Faculty

Answer 2

Step-by-step explanation:

$log_{10}(1 + tanx) \\ \\ \frac{1}{1 + tanx} .............$