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Region Area Ascertainment 2 (Posted on 2024-01-27) Difficulty: 3 of 5
Determine the area of the region bound by:
* y= sin x
* y= cos x
* y= tan x
with 0<=x<=π/2

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution solution | Comment 1 of 6
Plotting the three curves shows the area sought is that of a "triangle" whose vertices are (0,0) where the tangent curve meets the sine curve; (pi/4,1/sqrt(2)) where the sine meets the cosine; and the point where the cosine matches the tangent, whose x value falls between the other two:

x = 2*arctan(1/2*(1 + sqrt(5) - sqrt(2*(1 + sqrt(5))))) 
   =~ 0.666239432492515
   
Between x = 0 and this above value of x we need the integral of the tangent minus that of the sine. Beyond that, to x = pi/4 we need the integral of the cosine minus that of the sine. Since the amount subtracted in both instances the subtractions can be combined into one subtraction.

The indefinite integral of the tan function is -ln(cos(x)). Evaluated at zero you get -ln(1) = 0.
Evaluated at the above-mentioned x value, 0.666239432492515, you get 0.240605912529802, making that the needed tan integral.

The indefinite integral of the cosine function is sin(x), which we need to evaluate from 0.666239432492515 to pi/4 or about 0.785398163397448:

sin(0.785398163397448) - sin(0.666239432492515) =    0.0890727924366526

The integral of the sine curve is -cos(x) evaluated from zero to pi/4, to be subtracted from the sum of the two previous integrals. Its value is -0.707106781186548 -(-1) =  0.292893218813452.

0.240605912529802 + 0.0890727924366526 - 0.292893218813452 =  0.0367854861530026

  Posted by Charlie on 2024-01-27 09:56:53
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