# Area under an inverse trigonometric function

• Krushnaraj Pandya
In summary, the student attempted to find the area bounded by arcsinx, arccosx, and the x axis by integrating inverse functions. The hint suggested that the student think of the equations y = \arccos x## and y = \arcsin x## as x-functions of y and that this would lead to the same simple integral as a ##dy## integration. The student was able to solve the problem in their head without ever drawing a graph.
Krushnaraj Pandya
Gold Member

## 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?

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)##.

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.

LCKurtz said:
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?)

Krushnaraj Pandya said:
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.

Krushnaraj Pandya said:

## 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?

Krushnaraj Pandya said:
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!

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

## 1. What is the definition of the area under an inverse trigonometric function?

The area under an inverse trigonometric function is the region bounded by the curve of the function, the x-axis, and the vertical lines that intersect the x-axis at the endpoints of the interval.

## 2. How is the area under an inverse trigonometric function calculated?

The area under an inverse trigonometric function can be calculated using integration techniques. The integral of the function from the lower bound to the upper bound gives the area under the curve. This can be done using various methods such as substitution, integration by parts, or trigonometric identities.

## 3. Can the area under an inverse trigonometric function be negative?

Yes, the area under an inverse trigonometric function can be negative. This occurs when the function lies below the x-axis within the given interval. In this case, the value of the integral would be negative, indicating a negative area.

## 4. What is the significance of calculating the area under an inverse trigonometric function?

Calculating the area under an inverse trigonometric function can be useful in determining the average value of the function over a given interval. It can also help in finding the area of irregular shapes, as these can often be represented by inverse trigonometric functions.

## 5. Are there any real-world applications of finding the area under an inverse trigonometric function?

Yes, there are several real-world applications of finding the area under an inverse trigonometric function, such as calculating the work done by a variable force, finding the center of mass of an irregular object, and estimating the area under a probability distribution curve in statistics.

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