Verified answer

1)

∫ arcsin(t) dt >> We'll be using formula ∫ u dv = v u - ∫ v du

let: arcsin(t) = u ⇒du = dt /√(1 - t²) and t = v ⇒dt = dv

∫ arcsin(t) dt = t arcsin(t) - ∫ t [ dt /√(1 - t²)]

= t arcsin(t) - ∫ ( t) dt /√(1 - t²) >> To compute the integral you can use substitution:

u = 1 - t² ⇒du = - 2tdt >> ∫ ( t) dt /√(1 - t²) becomes:

- 1/2 ∫du /√u = - 1/2[2√u + C] = - √u + C >> sub back for u =1 - t²

= t arcsin(t) - [ - √(1 - t²) + C]

= t arcsin(t) + √(1 - t²) + C

2)

∫ sin² x cos³ x dx =

∫ sin² x cos² x cos x dx =

∫ sin² x (1 - sin² x) cos x dx >>

Now, we can use substitution.

let u = sin x ⇒du = cos x dx

∫ sin² x (1 - sin² x) cos x dx >> becomes>> ∫ u² (1 - u²) du =

∫ (u² - u^4) du = (1/3)u³ - (1/5)u^5 + C

Substitute back for u = sin x and you'll have your final answer:

(1/3)sin³ x - (1/5)sin^5 x + C

Best Regards.

For the first one, let:

u = arcsin(t) ==> u = 1/√(1 - t^2) dt

dv = dt ==> v = t.

By integration by parts:

∫ arcsin(t) dt

=> uv - ∫ v du

= t*arcsin(t) - ∫ t/√(1 - t^2) dt

= t*arcsin(t) + 1/2 ∫ (-2t)/√(1 - t^2) dt

= t*arcsin(t) + 1/2 ∫ 1/√(1 - t^2) d(1 - t^2)

= t*arcsin(t) + √(1 - t^2) + C.

For the second one, save a cosine to obtain:

∫ [sin^2(x) * cos^3(x)] dx

= ∫ [sin^2(x) * cos^2(x)] [cos(x) dx]

= ∫ {sin^2(x) * [1 - sin^2(x)]} [cos(x) dx]

= ∫ [sin^2(x) - sin^4(x)] [cos(x) dx].

Letting u = sin(x) <==> du = cos(x) dx yields:

∫ [sin^2(x) - sin^4(x)] [cos(x) dx]

= ∫ (u^2 - u^4) du

= (1/3)u^3 - (1/5)u^5 + C.

Back-substituting u = sin(x) gives:

∫ [sin^2(x) * cos^3(x)] dx = (1/3)sin^3(x) - (1/5)sin^5(x) + C.

I hope this helps!

I got for int [ arcsin t dt ] = t arcsin t + ( 1 / 2) sqrt ( 1 - t^2 ) + C

the second one is even easier.

int [ sin^2 x cos^3 x dx ]

= int [ sin^2 x ( 1 - sin^2 x ) cos x dx ]

= int [ (sin^2 x - sin^4 x ) cos x dx ]

= ( 1 / 3)sin^3 x - (1 / 5)sin^5 x + C