integral of 1/((arcsin(x))^3sqrt(1-x^2))
1) Let's compute the integral: ∫1arcsin(x)3·1 -x2dx
For the math expression: 1arcsin(x)3·1 -x2, since (a^n)' = n*a'*a^(n - 1), we try differentiating arcsin(x)-3 + 1
2) Let's perform the differentiation: ddx(arcsin(x)-3 + 1)
ⓘ The exponent is computed before the power: -3 + 1 = -2
ddx(arcsin(x)-2)
Apply the differentiation rule: ddx(fn) = n·ddx(f)·fn -1 with f = arcsin(x)
ⓘ The derivative of x^n is n*x^(n-1): ddx(arcsin(x)-2) = -2·ddx(arcsin(x))·arcsin(x)-3
-2·ddx(arcsin(x))·arcsin(x)-3
ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base: -2·ddx(arcsin(x))·arcsin(x)-3 = -2·ddx(arcsin(x))arcsin(x)3
-2·ddx(arcsin(x))arcsin(x)3
ⓘ The derivative of arcsin(x) is 1/sqrt(1 - x^2): ddx(arcsin(x)) = 11 -x2
-2·11 -x2arcsin(x)3
ⓘ Multiplying a fraction by a number multiplies its numerator: 2·11 -x2 = 2·11 -x2 = 21 -x2
-21 -x2arcsin(x)3
ⓘ Dividing a fraction by arcsin(x)3 is multiplying its denominator by arcsin(x)3: -21 -x2arcsin(x)3 = -21 -x2·arcsin(x)3
-21 -x2·arcsin(x)3
The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:
ⓘ The derivative of -12·1arcsin(x)2 is 1arcsin(x)3·1 -x2, so -12·1arcsin(x)2 is an antiderivative of 1arcsin(x)3·1 -x2: ∫1arcsin(x)3·1 -x2dx = -12·1arcsin(x)2
-12·1arcsin(x)2
ⓘ To multiply fractions, multiply the numerators together and the denominators together: -12·1arcsin(x)2 = -1·12·arcsin(x)2 = -12·arcsin(x)2
-12·arcsin(x)2
▶Indefinite integrals are defined up to an additive constant C, so this yields: -12·arcsin(x)2 + C