Thread: Another inverse sine integral check

Hi, If anyone could check the solution I got to this problem, that'd be great.

Q: The integral of xsin^-1(x) dx

My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

The answer I got was:

1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


Thanks a lot for the help, I appreciate it.

Originally Posted by alysha27

Hi, If anyone could check the solution I got to this problem, that'd be great.

Q: The integral of xsin^-1(x) dx

My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

The answer I got was:

1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


Thanks a lot for the help, I appreciate it.

That is correct.

RonL

Originally Posted by alysha27

Hi, If anyone could check the solution I got to this problem, that'd be great.

Q: The integral of xsin^-1(x) dx

My solution: To solve, I first used integration by parts, and then with the integral left over by parts ( of x squared over 2 times the square root of one minus x squared) I used trigonometric substitution to solve the rest.

The answer I got was:

1/2*x^2*sin^-1(x) - 1/4*sin^-1(x) + 1/4*x*(1-x^2)^1/2 +C


Thanks a lot for the help, I appreciate it.

The easiest way is to take the derivative of the anti-derivative and see if it returns back the function.