Fun little integral

Calculus Level 2

∫ 0 1 x 2 − 1 x 2 + 1 d x = a − π b \large \int_{0}^{1}{\frac{x^{2}-1}{x^{2}+1}dx} = a - \frac \pi b

If the equation above holds true for real numbers a a and b b , find a + b a+b .


The answer is 3.

Gilliën Fransz by Gilliën Fransz , 23, Netherlands

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

I = ∫ 0 1 x 2 − 1 x 2 + 1 d x = ∫ 0 1 x 2 + 1 − 2 x 2 + 1 d x = ∫ 0 1 ( 1 − 2 x 2 + 1 ) d x = x − 2 tan ⁡ − 1 x ∣ 0 1 = 1 − π 2 \begin{aligned} I & = \int_0^1 \frac {x^2-1}{x^2+1} dx \\ & = \int_0^1 \frac {x^2+1-2}{x^2+1} dx \\ & = \int_0^1 \left(1-\frac 2{x^2+1}\right) dx \\ & = x - 2 \tan^{-1} x \ \bigg|_0^1 \\ & = 1 - \frac \pi 2 \end{aligned}

Therefore, a + b = 1 + 2 = 3 a+b=1+2=\boxed{3} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, how did 1/x^2 + 1 become tan inverse of x after integrating ?

Aman thegreat - 3 years, 2 months ago

Log in to reply

It's an standard identity which we can obtained by making trigonometric substitution of x = a tan ⁡ θ x =a \tan\theta and for this case a = 1 a =1 .

Naren Bhandari - 3 years, 2 months ago

y = tan ⁡ − 1 x ⟹ tan ⁡ y = x sec ⁡ 2 y d y d x = 1 d y d x = 1 sec ⁡ 2 y = 1 tan ⁡ 2 y + 1 = 1 x 2 + 1 ⟹ y = ∫ 1 x 2 + 1 d x tan ⁡ − 1 x = ∫ 1 x 2 + 1 d x \begin{aligned} y & = \tan^{-1}x \\ \implies \tan y & = x \\ \sec^2 y \frac {dy}{dx} & = 1 \\ \frac {dy}{dx} & = \frac 1{\sec^2 y} = \frac 1{\tan^2 y+1} = \frac 1{x^2+1} \\ \implies y & = \int \frac 1{x^2+1} dx \\ \tan^{-1} x & = \int \frac 1{x^2+1} dx \end{aligned}

Chew-Seong Cheong - 3 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...