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### Integration Over a Complicated Region

```
Date: 11/1/96 at 19:12:28
From: brenchi
Subject: Integration

The region R in the Cartesian plane consists of the points whose
coordinates simultaneously satisfy the five relations:

0 <= x <= (pi/2),  y >= 0,  y >= sinx,  y <= cosx,   y <= tanx

Prove that the area of R is:

(1/2)*{2*2^(1/2) - 1 - 5^(1/2) - log {[5^(1/2)-1]/2}} units square

From Billy
```

```
Date: 11/2/96 at 16:39:1
From: Doctor Anthony
Subject: Re: Integration

Given that the region R in the Cartesian plane consists of the
points whose coordinates simultaneously satisfy the five relations:

0 <= x <= (pi/2),  y >= 0,  y >= sinx,  y <= cosx,  y <= tanx,

sketch the graphs and note that if A is the point where the
tan(x) and cos(x) curves cut, we need the areas:

INT(0 to A)[tan(x)-sin(x)]dx  +  INT(A to pi/4)[cos(x)-sin(x)]dx

The point A has tan(x) = cos(x)   so  sin(x) = cos^2(x)

sin(x) = 1 - sin^2(x)       sin^2(x) + sin(x) - 1 = 0

Using the quadratic formula,  sin(x) = [sqrt(5)-1]/2

x = arcsin{[sqrt(5)-1]/2}

This is the x coordinate of point A, where cos(x) = sqrt(sin(x)).

Integrating the expressions given above:

[-ln(cos(x)) + cos(x) from 0 to A] + [sin(x) + cos(x) from A to pi/4]

= [-ln(sqrt(sin(x)))+sqrt(sin(x)-0-1] + [1/sqrt(2) + 1/sqrt(2) -
sin(x)-sqrt(sin(x)]    where sin(x) = [sqrt(5)-1]/2

= [-(1/2)ln{(sqrt(5)-1)/2} -1 + 2/sqrt(2) - (sqrt(5)-1)/2

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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