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Area under an inverse trigonometric function

  • #1
Krushnaraj Pandya
Gold Member
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71

Homework Statement


Find the area bounded by arcsinx, arccosx and the x axis.
Hint-you don't need to integrate arcsinx and arccosx

Homework Equations


All pertaining to calculus

The Attempt at a Solution


I drew the correct graph and marked their intersection at (1/√2, pi/4) and painstakingly found the answer by integrating inverse functions as (root 2 - 1).
Any insight on how to use the hint and make this easier?
 

Answers and Replies

  • #2
Stephen Tashi
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One thought is to consider the rectangle with vertices ##(0,0),(1,0),(1,\frac{\pi}{2}),(0,\frac{\pi}{2})## and find other areas within that rectangle by integrating ##\sin(y)## or ##\cos(y)##.
 
  • #3
LCKurtz
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Or draw the graphs of ##y=\sin x## and ##y = \cos x## and look for a congruent area. It will give you a trivial integration.
 
  • #4
Krushnaraj Pandya
Gold Member
697
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Or draw the graphs of ##y=\sin x## and ##y = \cos x## and look for a congruent area. It will give you a trivial integration.
Elegant! yet I had to use a grapher to see the congruency clearly. Is there another way I can see this easily e.g during a exam (perhaps an algebraic method?)
 
  • #5
LCKurtz
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Elegant! yet I had to use a grapher to see the congruency clearly. Is there another way I can see this easily e.g during a exam (perhaps an algebraic method?)
I suggested that because most people find it easy to think of ##y## in terms of ##x##. But if you take your equations ##y = \arccos x## and ##y = \arcsin x## and think of them as ##x## as a function of ##y## you have ##x = \cos y## and ##x = \sin y## and the calculation of the area is the same simple integral as a ##dy## integration.
 
  • #6
Ray Vickson
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Homework Statement


Find the area bounded by arcsinx, arccosx and the x axis.
Hint-you don't need to integrate arcsinx and arccosx

Homework Equations


All pertaining to calculus

The Attempt at a Solution


I drew the correct graph and marked their intersection at (1/√2, pi/4) and painstakingly found the answer by integrating inverse functions as (root 2 - 1).
Any insight on how to use the hint and make this easier?
Elegant! yet I had to use a grapher to see the congruency clearly. Is there another way I can see this easily e.g during a exam (perhaps an algebraic method?)
Think of it this way: draw the graphs of ##y = \sin x## and ##y = \cos x## for ##0 \leq x \leq \pi/2##. Here, the x-axis is horizontal and the y-axis is vertical Now rotate the graph paper through 90 degrees, so the old x-axis is now vertical and the old y-axis is now horizontal. You would now be looking at the plots of ##\arcsin y## and ##\arccos y##, ##0 \leq y \leq 1.## On the rotated graph, shade in the required area between the plots of ##\arcsin y, \arccos y## and ##y = 0##. Now rotate the graph paper back to its original orientation, with the x-axis horizontal and the y-axis vertical again. All you will have done is rotated an area through 90 degrees, without changing its numerical value. That means that you can evaluate the area by looking at plots of ##\sin x## and ##\cos x,## which might---and in this case, does---lead to an easier problem.

Here comes the final trick: you can do all that in your head, without ever drawing a single actual graph!
 
  • #7
Krushnaraj Pandya
Gold Member
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Thank you very much everyone! All the inversion and the change in function (in terms of y instead of x) seemed a bit strange but I finally wrapped my head around it after solving this multiple times in my head, its quite elegant in fact. Thank you for your help :D
 

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