Solving Antiderivative of arcsin(x^0.5)/(x^0.5)

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SUMMARY

The discussion focuses on solving the antiderivative of the function arcsin(x^0.5)/(x^0.5). The recommended approach involves substituting u = x^0.5, which transforms the integral into arcsin(y) dy. Following this substitution, integration by parts is applied, where u = arcsin(y), du = sqrt(1-y^2) dy, dv = dy, and v = y. This method simplifies the integral, making it manageable to solve.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with trigonometric functions, particularly arcsin.
  • Knowledge of substitution methods in calculus.
  • Basic algebra skills for manipulating expressions.
NEXT STEPS
  • Study the process of integration by parts in calculus.
  • Learn about the properties and applications of the arcsin function.
  • Explore substitution methods for simplifying integrals.
  • Practice solving integrals involving square roots and trigonometric functions.
USEFUL FOR

Students and educators in calculus, mathematicians focusing on integration techniques, and anyone looking to deepen their understanding of antiderivatives involving trigonometric functions.

okhjonas
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Hello, can somebody here tell me how to solwe the antiderivate of

arcsin(x^0,5)/(x^0,5)

I know the answer but not the solution..
 
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okhjonas said:
Hello, can somebody here tell me how to solwe the antiderivate of

arcsin(x^0,5)/(x^0,5)

I know the answer but not the solution..

let u = x^{.5}
 
First make a substiution, so that you are integrating arcsin(y)dy. Then integrate by parts u = arcsin(y) du = sqrt(1-y^2)dy dv=dy v =y. The resulting integral of vdu shouldn't be too hard to solve.
 

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