Integrating inverse sin3x?

Question

Hi,

There is a question that's bugging me and its finding the integral of inverse sin3x using partial fractions(dummy functions).

I am not sure if i am right but i got :

x(inverse)sin3x + 1/9 square root 1-9x^2 + C

Thanks for your help in advanced.

Answer 1

integrate by parts:



u = arcsin(3x) . . . . . . . . . dv = dx

du = 3/√(1-9x²) dx . . . . . . v = x



∫ arcsin(3x) dx = x arcsin(3x) - ∫ 3x/√(1-9x²) dx



Integrate by substitution:

t = 1 - 9x²

dt = -18x dx

3x dx = -1/6 dt



∫ arcsin(3x) dx = x arcsin(3x) - ∫ 1/√(1-9x²) * 3x dx

..................... = x arcsin(3x) - ∫ 1/√t * -1/6 dt

..................... = x arcsin(3x) + 1/6 ∫ 1/√t dt

..................... = x arcsin(3x) + 1/6 * 2√t + C

..................... = x arcsin(3x) + 1/3 √(1-9x²) + C

Answer 2

∫ sin^(-1)(3x) dx



By substitution method, let u = 3x and du = 3 dx.



= (1/3) ∫ sin^(-1)(u) du



By integration by parts, ∫ f(dg) = f(g) - ∫ g(df):



f = sin^(-1)(u)

dg = du

g = u

df = 1/√(1 - u²) du



Putting that altogether, we have:



= (1/3)[usin^(-1)(u) - ∫ u/√(1 - u²) du]



By substitution method, let s = 1 - u² and ds = -2u du:



= (1/3)[usin^(-1)(u) - (½) ∫ 1/√s ds]

= (1/3)[usin^(-1)(u) - (½) ∫ s^(-½) ds]



By integral power rule, ∫ xⁿ dx = x^(n + 1)/(n + 1), the integral of s^(-½) is s^(-½ + 1)/(-½ + 1).



= (1/3)[usin^(-1)(u) - (½)s^(-½ + 1)/(-½ + 1) + c]

= (1/3)[usin^(-1)(u) - (½)s^(½)/(½) + c]

= (1/3)[usin^(-1)(u) - s^(½) + c]



Substitute s with 1 - u² and u with 3x:



= (1/3)[3xsin^(-1)(3x) - √(1 - u²) + c]

= (1/3)[3xsin^(-1)(3x) - √(1 - (3x)²) + c]

= (1/3)[3xsin^(-1)(3x) - √(1 - 9x²) + c]

= xsin^(-1)(3x) - √(1 - 9x²)/3 + c



Hence, the integral of the given expression is xsin^(-1)(3x) - √(1 - 9x²)/3 + c.



You are correct for that problem! Nice job! Keep up the good work!

Answer 3

(1/3) sqrt(1-9x^2) + x sin^-1 3x + c

Answer 4

∫(arcsin(3x)) dx [here arcsin means sine inverse]



Let y = arcsin(3x), sin(y) = 3x, cos(y) dy = 3 dx.



= (1/3) ∫(y cos(y)) dy



Now integrating by parts



= (1/3) (y ∫cos(y) dy - ∫(∫cos(y) dy) dy)



= (1/3) (y sin(y) - ∫sin(y) dy)



= (1/3) (y sin(y) + cos(y))



= (1/3) (arcsin(3x) sin(arcsin(3x)) + cos(arcsin(3x)))



= (1/3) (3x arcsin(3x) + √(1 - 9x^2))



= x arcsin(3x) + (1/3)√(1 - 9x^2)