Solving the Integral of Arcsin: Techniques and Variations

Understanding how to solve the integral of arcsin is quite valuable in advanced calculus and real-world applications. This article explores various methods, including variable substitution and integration by parts, to find the integral of arcsin x. We'll walk through the process step-by-step and demonstrate how to apply these techniques effectively.

Method 1: Integration by Parts

The integral of arcsin x can be solved using the integration by parts formula. This method relies on the relationship between arcsin x and sin y. Let's delve into the process.

Step 1: Set up the integral and apply the integration by parts formula.

Given: ( y text{arcsin } x ) Therefore, ( x sin y )

Now, use the integration by parts formula:

( int y , dx x y - int x , dy )

Substitute and simplify:

( int text{arcsin } x , dx x , text{arcsin } x - int sin y , dy )

Solve the integral on the right:

( int text{arcsin } x , dx x , text{arcsin } x cos y C )

Substitute ( y text{arcsin } x ) and ( cos y sqrt{1 - x^2} ):

( int text{arcsin } x , dx x , text{arcsin } x , sqrt{1 - x^2} C )

Method 2: Variable Substitution

Another approach to solving the integral of arcsin x is through variable substitution. This method simplifies the integral by transforming the variable.

Step 1: Set up the integral with a substitution.

Let ( x sin t ). Then, ( dx cos t , dt ).

Substitute into the integral:

( int text{arcsin } x , dx int t , cos t , dt )

Step 2: Use integration by parts for the new integral.

( u t, , dv cos t , dt ), so ( du dt, , v sin t )

Apply integration by parts:

( int t , cos t , dt t , sin t - int sin t , dt )

Simplify the result:

( int t , cos t , dt t , sin t cos t C )

Substitute back ( t text{arcsin } x ) and ( cos t sqrt{1 - x^2} ):

( int text{arcsin } x , dx text{arcsin } x , sqrt{1 - x^2} sqrt{1 - x^2} C )

Combine terms:

( int text{arcsin } x , dx text{arcsin } x , sqrt{1 - x^2} C )

Conclusion

Both methods effectively solve the integral of arcsin x. Understanding these techniques not only provides a deeper grasp of calculus but also enhances problem-solving skills in mathematics and related fields. Whether you prefer the direct approach of integration by parts or the elegance of variable substitution, both methods yield the same result, simplifying the integral in a straightforward manner.

For further exploration, you can apply these methods to more complex integrals involving inverse trigonometric functions. Additionally, practice with different functions can reinforce your understanding and improve your ability to tackle similar problems efficiently.