This is solved by using substitution:
Let x = sin θ
→ dx = cos θ dθ
and √(1 - x²) = √(1 - sin² θ) = cos θ
and θ = arcsin x
→ ∫ (x + 1)/√(1 - x²) dx
= ∫ ((sin θ + 1)/ cos θ) cos θ dθ
= ∫ sin θ + 1 dθ
= -cos θ + θ + b
= θ - cos θ + b
= arcsin x - √(1 - x²) + c
replace square root o x with t.
square root x
What square root property is essential to solve any radical equation involving square root?
The square root of 28 is an irrational number that can't be expressed as a fraction
square
The square root of 2 is irrational, and therefore, this equation is unsolvable.
the integral of the square-root of (x-1)2 = x2/2 - x + C
If you move that 46x over the equals, and if you square both sides, you can get rid of that square root, and do the equation normally. do the math
replace square root o x with t.
square root x
What square root property is essential to solve any radical equation involving square root?
The square root of 28 is an irrational number that can't be expressed as a fraction
The square root is not integral : about 7585.998 (round to 7586 - this square is 54547396, so this could be a transposition error).
square
The square root of 81 is an integral number.
-cos(x) + constant
Technically,no. A radical equation has a radical (Square root) in it, and has two solutions because the square root can be positive or negative.