A bit off tangent (My sixteenth integration problem)

Calculus Level 3

$\large \displaystyle \int_{0}^{1} \dfrac{1-x^{2}}{1+x^{2}} \, dx$

If the above integral can be expressed in the form

$\dfrac{a \pi}{b} - c$

where $a, b, c$ are positive integers and $a, b$ are coprime, find $a+b+c$ .

The answer is 4.

1 solution

Rishabh Jain
Feb 23, 2016

$\large\mathfrak{K}= \displaystyle \int_{0}^{1} \dfrac{1-x^{2}}{1+x^{2}} \, dx$ $\large =\int_{0}^1 (\dfrac{2}{1+x^2}-1 )\, dx$ $=2\tan^{-1}x-x|_{0}^1$ $\large =\dfrac{\pi}{2}-1$ $\huge \therefore 1+2+1=\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{\textbf{4}}}}}}$

Nice! I did it with a trigonometric substitution but I find your way to be much faster! :)

Hobart Pao - 5 years, 3 months ago

$\huge \color{#302B94}{\mathbb{THANKS}}$

Rishabh Jain - 5 years, 3 months ago

A bit off tangent (My sixteenth integration problem)

The answer is 4.

1 solution

0 pending reports