Evaluate-0-1-tan-1-x-x-dx-

Question Number 12101 by Nayon last updated on 13/Apr/17

E v a l u a t e \int_{0}^{1} \frac{{t a n}^{- 1} (x)}{x} d x

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Apr/17

x = t g φ \Rightarrow d x = \frac{d φ}{1 + {t g}^{2} φ} (\begin{matrix} x = 1 \Rightarrow φ = \frac{π}{4} \\ x = 0 \Rightarrow φ = 0 \end{matrix})

I = \int \frac{{t g}^{- 1} (t g φ)}{t g φ} . \frac{d φ}{1 + {t g}^{2} φ} = \int \frac{φ d φ}{t g φ (1 + {t g}^{2} φ)}

= \frac{φ}{t g φ} - \int \frac{d φ}{t g φ} = \frac{φ}{t g φ} - \int \frac{c o s φ}{s i n φ} d φ =

\frac{φ}{t g φ} - l n ∣ s i n φ ∣ + C .

I = \frac{π}{4 \times 1} - l n (\frac{\sqrt{2}}{2}) = \frac{π}{4} + \frac{1}{2} l n 2 .

(I = \frac{{t g}^{- 1} x}{x} - l n ∣ \frac{x}{\sqrt{1 + x^{2}}} ∣ + C) .