Understanding the Derivative of Inverse Trig Functions

Click For Summary
SUMMARY

The discussion focuses on understanding the derivative of inverse trigonometric functions, specifically the arcsine function. The key takeaway is that when differentiating arcsine of (u/a), a substitution is necessary to simplify the integrand to the form 1/√(1-x²). This substitution involves replacing u/a with x, which effectively cancels out the factor of 1/a, leading to a clean result in the integration process. Participants confirmed their understanding through collaborative explanation and clarification of the steps involved.

PREREQUISITES
  • Understanding of inverse trigonometric functions, particularly arcsine.
  • Knowledge of basic calculus, including differentiation and integration techniques.
  • Familiarity with variable substitution in integration.
  • Proficiency in manipulating algebraic expressions during calculus operations.
NEXT STEPS
  • Study the properties and derivatives of inverse trigonometric functions.
  • Learn about variable substitution techniques in integration.
  • Explore the application of the Fundamental Theorem of Calculus.
  • Practice problems involving the integration of functions with inverse trigonometric components.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone seeking to deepen their understanding of inverse trigonometric functions and their derivatives.

alijan kk
Messages
130
Reaction score
5

Homework Statement


question.PNG

why this formula works ?

Homework Equations

The Attempt at a Solution


when i take the derivative of the right side ,,, there is an additional "a" in the numerator in place of 1,, why the derivative of arcsine of (u/a) not exactly same with the expression under the integral sign
28217670_236535466916604_1075396015_o.jpg
 

Attachments

  • question.PNG
    question.PNG
    5.6 KB · Views: 698
  • 28217670_236535466916604_1075396015_o.jpg
    28217670_236535466916604_1075396015_o.jpg
    24.7 KB · Views: 339
Last edited:
Physics news on Phys.org
In order to finally perform the integration in the last step you need to do a substitution, say replacing ##u/a## by ##x## so that your integrand has the form ##1/\sqrt{1-x^2}##. In order to do that you need to make the other changes that are part of doing a variable substitution in integration. Those changes should cancel out the factor ##1/a## and leave you with the nice, clean result you want.
 
  • Like
Likes   Reactions: alijan kk
now i understand it very well,,,, thankyou for they reply
andrewkirk said:
In order to finally perform the integration in the last step you need to do a substitution, say replacing ##u/a## by ##x## so that your integrand has the form ##1/\sqrt{1-x^2}##. In order to do that you need to make the other changes that are part of doing a variable substitution in integration. Those changes should cancel out the factor ##1/a## and leave you with the nice, clean result
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
16
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K