I need help with Ito's Integral?

1)

You are using Bs to denote one letter. I have assumed T is the upper limit and 0 is the lower limit.

Let u= Bs

du = dBs

∫ sin(Bs) dBs = ∫ sin(u) du = -cos(u)

= -cos(Bs)

Let F(Bs) = -cos(Bs)

substitute the upper limit T

F(T) = -cos(T)

substitute the lower limit 0

F(0) = -cos(0) = -1

F(T)-F(0) = -cos(T) - (-1) = 1 -cos(T)

2)

I have assumed T is the upper limit and 1 is the lower limit

Write this as:

∫ dx / (x^2+x^4) where x= Bs

= ∫ dx / x^2(1+x^2)

decompose into partial fractions

1/ x^2(1+x^2) = A/x + B/x^2 + (Cx+D)/(1+x^2)

multiply both sides by x^2(1+x^2)

1 = Ax (1+x^2) + B(1+x^2) + (Cx+D)x^2

Equate the coefficient of x^3 from both sides

0 = A + C

A = -C

Equate the coefficient of x^2 from both sides

0 = B+D

B = -D

Equate the coefficient of x from both sides

0 = A

C = 0

Equate the constant terms from both sides

1 = B

D = 1

A=0; B=1; C=0; D=1

1/ x^2(1+x^2) = A/x + B/x^2 + (Cx+D)/(1+x^2)

∫ 1/ x^2(1+x^2) dx = 1 ∫ dx /x^2 + 1∫ dx /(1+x^2)

= -1/x + tan^-1(x)

replace x by Bs

= -1/ Bs + tan^-1( Bs)

∫ dx / (Bs^2+Bs^4) = - 1/Bs + tan^-1( Bs)

Let F(Bs) = - 1/Bs + tan^-1( Bs)

substitute the upper limit T

F(T) = -1/ T + tan^-1(T)

substitute the lower limit 1

F(1) = -1 + tan^-1(1)

F(T)-F(1) = -1/ T + tan^-1(T) +1 - tan^-1(1)

= -1/ T + tan^-1(T) +1 - pi/4

I'm assuming Bs is a Brownian motion with B0 = 0. More information about Bs is needed for computing the indefinite integral in the second part.

I'll calculate the first integral. Let f(x) = -cos x. By Ito's formula,

f(BT) - f(B0) = ∫[0,T] f'(Bs) dBs + 1/2 ∫[0,T] f"(Bs) ds,

1 - cos BT = ∫[0,T] sin Bs dBs + 1/2 ∫[0,T] cos Bs ds.

Therefore

∫[0,T] sin Bs dBs

= 1 - cos BT - (1/2) ∫[0,T] cos Bs ds.

I'm gonna call this a different variable

Bs = x

sin(x) * dx

The integral of that is -cos(x) + C

-cos(0) - (-cos(T)) =>

-1 + cos(T) =>

cos(T) - 1

dt / (t^2 + t^4) =>

dt / (t^2 * (1 + t^2))

A/t + B/t^2 + (Ct + D) / (1 + t^2)

A * t * (1 + t^2) + B * (1 + t^2) + (Ct + D) * t^2 = 1

At^3 + At + Bt^2 + B + Ct^3 + Dt^2 = 0t^3 + 0t^2 + 0t + 1

A + C = 0

B + D = 0

A = 0

B = 1

A + C = 0

0 + C = 0

C = 0

B + D = 0

1 + D = 0

D = -1

dt/t^2 - dt/(1 + t^2)

tan(u) = t

sec(u)^2 * du = dt

dt / t^2 - sec(u)^2 * du / (1 + tan(u)^2)

dt / t^2 - sec(u)^2 * du / sec(u)^2 =>

dt / t^2 - du

Integrate

-1/t - u + C =>

-1/t - arctan(t) + C

From T to 1

-1/1 - arctan(1) + 1/T + arctan(T)

-1 - pi/4 + 1/T + arctan(T)

1/T + arctan(T) - (4 + pi) / 4

1.)

Integrates to -cos(Bs), and evaluating this from T to 0, you get -1 + cos(T).

2.)

Re-write 1/(Bs^2 + Bs^4) as 1/(Bs^2(1 + Bs^2)) = (Bs^2 + 1 - Bs^2)/(Bs^2(1 + Bs^2))

= 1/(Bs^2) - 1/(1 + Bs^2)

Integrating, we get -1/(Bs) - arctan(Bs) eval. from T to 1

= -1 - pi/4 + 1/T + arctan(T)