What's with all the trigonometry?

Calculus Level 4

∫ 0 1 ∫ arcsin ⁡ ( y ) π / 2 cos ⁡ ( x ) e − cos ⁡ 2 ( x ) d x d y = a b ( 1 − c e ) \large \int_{0}^{1} \int_{\arcsin{(y)}}^{\pi/2} \cos{(x)}e^{-\cos^2{(x)}} dx \ dy = \dfrac{a}{b} \left( 1-\dfrac{c}{e} \right)

Given that a , b , c ∈ Z + a,b,c \in \mathbb{Z^+} and b b is a prime number, find a + b + c a+b+c .

Notation: e ≈ 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is 4.

Akeel Howell by Akeel Howell , 22, USA

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.

2 solutions

First Last
Nov 11, 2017

Following Reynan's hint change the order of integration by looking at the region of integration

Changing the double integral from I = ∫ 0 1 ∫ arcsin ⁡ ( y ) π 2 cos ⁡ ( x ) e − cos ⁡ 2 ( x ) d x d y \displaystyle\textbf{I}=\int_0^1\int_{\arcsin{(y)}}^\frac{\pi}{2}\cos{(x)}e^{-\cos^2{(x)}}dxdy to ∫ 0 π 2 ∫ 0 sin ⁡ ( x ) cos ⁡ ( x ) e − cos ⁡ 2 ( x ) d y d x \displaystyle\int_0^\frac{\pi}{2} \int_0^{\sin{(x)}}\cos{(x)}e^{-\cos^2{(x)}}dydx

and noticing that − cos ⁡ 2 ( x ) = sin ⁡ 2 ( x ) − 1 -\cos^2{(x)} = \sin^2{(x)}-1 :

I = 1 2 ∫ 0 π 2 2 sin ⁡ ( x ) cos ⁡ ( x ) e sin ⁡ 2 ( x ) − 1 d x set u = s i n 2 ( x ) \displaystyle\textbf{I}=\frac1{2}\int_0^\frac{\pi}{2}2\sin{(x)}\cos{(x)}e^{\sin^2{(x)}-1}dx\quad\quad\text{set } u = sin^2{(x)}

I = 1 2 ∫ 0 1 e u − 1 d u = 1 2 ( 1 − 1 e ) \displaystyle\textbf{I}=\frac1{2}\int_0^1e^{u-1}du = \frac1{2}(1-\frac1{e})

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Reynan Henry
Nov 11, 2017

Hint: Try swapping x and y

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...