Find the Integral

Calculus Level 4

∫ 0 π / 4 tan ⁡ − 1 ( 1 − cos ⁡ 2 x 1 + cos ⁡ 2 x ) d x = ? \large \int_0^{\pi/4}\tan^{-1}\left( \sqrt{\frac {1-\cos2x}{1+\cos2x}} \right) \ dx = ?

The closed form of the integral above can be expressed as π a b \dfrac {\pi^a}b , where a a and b b are positive integers. Find a + b a+b .


The answer is 34.

Md Zuhair by Md Zuhair , 20, India

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

Chew-Seong Cheong
Oct 20, 2016

I = ∫ 0 π / 4 tan ⁡ − 1 ( 1 − cos ⁡ 2 x 1 + cos ⁡ 2 x ) d x = ∫ 0 π / 4 tan ⁡ − 1 ( 1 − 2 cos ⁡ 2 x + 1 1 + 2 cos ⁡ 2 x − 1 ) d x = ∫ 0 π / 4 tan ⁡ − 1 ( 2 − 2 cos ⁡ 2 x 2 cos ⁡ 2 x ) d x = ∫ 0 π / 4 tan ⁡ − 1 ( sin ⁡ 2 x cos ⁡ 2 x ) d x = ∫ 0 π / 4 tan ⁡ − 1 ( tan ⁡ 2 x ) d x = ∫ 0 π / 4 tan ⁡ − 1 ( tan ⁡ x ) d x = ∫ 0 π / 4 x d x = x 2 2 ∣ 0 π / 4 = π 2 32 \begin{aligned} I & = \int_0^{\pi/4} \tan^{-1} \left( \sqrt{\frac {1-\cos 2x}{1+\cos 2x}} \right) dx \\ & = \int_0^{\pi/4} \tan^{-1} \left( \sqrt{\frac {1-2\cos^2 x + 1}{1+2\cos^2x -1}} \right) dx \\ & = \int_0^{\pi/4} \tan^{-1} \left( \sqrt{\frac {2-2\cos^2x}{2\cos^2x}} \right) dx \\ & = \int_0^{\pi/4} \tan^{-1} \left( \sqrt{\frac {\sin^2x}{\cos^2x}} \right) dx \\ & = \int_0^{\pi/4} \tan^{-1} \left( \sqrt{\tan^2 x} \right) dx \\ & = \int_0^{\pi/4} \tan^{-1} \left( \tan x \right) dx \\ & = \int_0^{\pi/4} x \ dx = \frac {x^2}2 \ \bigg|_0^{\pi/4} = \frac {\pi^2}{32} \end{aligned}

⟹ a + b = 2 + 32 = 34 \implies a+b = 2+32 = \boxed{34}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...