integration with tanx

Calculus Level 3

∫ 0 1 tan ⁡ x x d x = ? \int_0^1 \frac {\tan x}x dx = \ ?


The answer is 1.1491.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Chew-Seong Cheong
Jul 17, 2020

One way to solve this is using Maclaurin series as follows:

I = ∫ 0 1 tan ⁡ x x d x = ∫ 0 1 1 x ( x + 1 3 x 3 + 2 15 x 5 + 17 315 x 7 + 62 2835 x 9 + ⋯ ) d x = ∫ 0 1 ( 1 + 1 3 x 2 + 2 15 x 4 + 17 315 x 6 + 62 2835 x 8 + ⋯ ) d x = [ x + 1 3 ⋅ 3 x 3 + 2 15 ⋅ 5 x 5 + 17 315 ⋅ 7 x 7 + 62 2835 ⋅ 9 x 9 + ⋯ ] 0 1 = 1 + 1 3 ⋅ 3 + 2 15 ⋅ 5 + 17 315 ⋅ 7 + 62 2835 ⋅ 9 + ⋯ ≈ 1.15 \begin{aligned} I & = \int_0^1 \frac {\tan x}x dx \\ & = \int_0^1 \frac 1x \left(x + \frac 13x^3 + \frac 2{15}x^5 + \frac {17}{315}x^7 + \frac {62}{2835}x^9 + \cdots \right) dx \\ & = \int_0^1 \left(1 + \frac 13x^2 + \frac 2{15}x^4 + \frac {17}{315}x^6 + \frac {62}{2835}x^8 + \cdots \right) dx \\ & = \left[x + \frac 1{3\cdot3} x^3 + \frac 2{15\cdot 5}x^5 + \frac {17}{315 \cdot 7}x^7 + \frac {62}{2835 \cdot 9}x^9 + \cdots \right]_0^1 \\ & = 1 + \frac 1{3\cdot3} + \frac 2{15\cdot 5} + \frac {17}{315 \cdot 7} + \frac {62}{2835 \cdot 9} + \cdots \\ & \approx \boxed{1.15} \end{aligned}

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