Problem to solve
Find the indefinite integral \[\int\arcsin^4[x] dx\]
Solution
STEP 1
Take the integral: integral sin^(-1)(x)^4 dx
STEP 2
For the integrand sin^(-1)(x)^2, integrate by parts, integral f dg = f g - integral g df, where f = sin^(-1)(x)^4, dg = dx, df = (4 sin^(-1)(x)^3)/sqrt(1 - x^2) dx, g = x: = x sin^(-1)(x)^4 - 4 integral(x sin^(-1)(x)^3)/sqrt(1 - x^2) dx
STEP 3
For the integrand (x sin^(-1)(x)^3)/sqrt(1 - x^2), integrate by parts, integral f dg = f g - integral g df, where f = sin^(-1)(x)^3, dg = x/sqrt(1 - x^2) dx, df = (3 sin^(-1)(x)^2)/sqrt(1 - x^2) dx, g = -sqrt(1 - x^2): = 4 sqrt(1 - x^2) sin^(-1)(x)^3 + x sin^(-1)(x)^4 - 12 integral sin^(-1)(x)^2 dx
STEP 4
For the integrand sin^(-1)(x)^2, integrate by parts, integral f dg = f g - integral g df, where f = sin^(-1)(x)^2, dg = dx, df = (2 sin^(-1)(x))/sqrt(1 - x^2) dx, g = x: = -12 x sin^(-1)(x)^2 + 4 sqrt(1 - x^2) sin^(-1)(x)^3 + x sin^(-1)(x)^4 + 24 integral(x sin^(-1)(x))/sqrt(1 - x^2) dx
STEP 5
For the integrand (x sin^(-1)(x))/sqrt(1 - x^2), integrate by parts, integral f dg = f g - integral g df, where f = sin^(-1)(x), dg = x/sqrt(1 - x^2) dx, df = 1/sqrt(1 - x^2) dx, g = -sqrt(1 - x^2): = -24 sqrt(1 - x^2) sin^(-1)(x) - 12 x sin^(-1)(x)^2 + 4 sqrt(1 - x^2) sin^(-1)(x)^3 + x sin^(-1)(x)^4 + 24 integral1 dx
STEP 6
The integral of 1 is x: Answer: | | = 4 sqrt(1 - x^2) sin^(-1)(x)^3 - 24 sqrt(1 - x^2) sin^(-1)(x) + 24 x + x sin^(-1)(x)^4 - 12 x sin^(-1)(x)^2 + constant
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Number
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Edited by tododemate® - 2021