Given λ > 0 compute the expectation of the following expression

Question

Given $\lambda > 0$, compute $\mathbb{E}[e^{-\lambda B(t)}\int_{0}^{t}e^{\lambda B(s)}dB(s)]$

My approach so far,

Let $$X(t) = e^{\lambda B(t)}; X(0)=1,$$ $$dX(t) = X(t)dB(t) + \frac{1}{2}X(t)dt$$ $$X(t) = 1 + \int_{0}^{t}X(s)dB(s) + \frac{1}{2}\int_{0}^{t}X(s)ds$$ $$\int_{0}^{t}X(s)dB(s) = X(t) - 1 - \frac{1}{2}\int_{0}^{t}X(s)ds$$

So, $$\mathbb{E}\Big[e^{-\lambda B(t)}\int_{0}^{t}e^{\lambda B(s)}dB(s)\Big] = \mathbb{E}\Big[e^{-\lambda B(t)}\Big[e^{\lambda B(t)}-1-\frac{1}{2}\int_{0}^{t}e^{\lambda B(s)}ds\Big]\Big] = \mathbb{E}\Big[1 - e^{-\lambda B(t)} - e^{-\lambda B(t)}\Big(\frac{1}{2}\int_{0}^{t}e^{\lambda B(s)}ds\Big)\Big].$$

How do I go from here? More specifically, how do I calculate $\mathbb{E}\Big[\frac{1}{2}\int_{0}^{t}e^{\lambda B(s)}ds\Big]$?

Edited after seeing the answers: $$\mathbb{E}\Big[1 - e^{-\lambda B(t)} - \frac{1}{2}\int_{0}^{t}e^{\lambda B(s)}ds\Big] = 1 - e^{\lambda^2t/2} - (e^{\lambda^2t/2}-1)/\lambda^2$$

Here, I have used the fact that $M_{B(t)}(\lambda)= M_{B(t)}(-\lambda)$, where, $M_{B(t)}(\lambda)$ is the Moment Generating function of B(t) $\sim N(0,t)$.

Answer 1

The last equality is not true, it should be $$ \mathbb E[1-e^{-\lambda B_t}- \frac{e^{-\lambda B_t}}{2}\int_0^t e^{\lambda B_s}ds] = 1 - e^{\frac{\lambda^2 t}2} - \frac{1}{2}\int_0^t e^{\frac{\lambda^2(t-s)}{2}}ds = 1-e^{\lambda^2t/2} - \frac{1}{\lambda^2}(e^{\lambda^2 t/2}-1) $$

Answer 2

Hint

Apply Fubini yields $$\mathbb E\left[\int_0^t e^{\lambda B_s}\,\mathrm d s\right]=\int_0^t\mathbb E[e^{\lambda B_s}]\,\mathrm d s$$

Answer 3

$E\int_0^{t}e^{\lambda B(s)}ds=\int_0^{t}Ee^{\lambda B(s)}ds=\int_0^{t}e^{\lambda^{2} s/2} ds=\frac 2 {\lambda^{2}}(e^{\lambda^{2} t/2}-1) $.