Here is the exercise I have been struggling to solve. It is taken from this book by Revuz and Yor: link. Here is the full text of the problem ( Exercise 3.32, chapter 4).
Exercise (3.32). Let $B$ and $C$ be two independent Brownian Motions started at $0$. Prove that
$$ \int _0 ^1 (B_t + C_{1-t})^2 dt \,{\buildrel (d) \over =}\, \int _0 ^1 (B_t ^2 + (B_1 - B_t)^2) dt $$
[Hint: The Laplace transform in $(\frac{\lambda ^2 }{2})$ of the right-hand side is the characteristic function in $\lambda$ of the sum of two stochastic integrals; see Exercise (2.19)]
I have been trying to apply formula $E \exp(\lambda M_t) = E \exp (\frac{\lambda ^2}{2} \langle M , M \rangle _t)$ for a continiouos martingale $M$, but I wasn't able to come up with a good idea what should be $M$. For example, we may write
$$ E \exp\left(\frac{\lambda ^2 }{2} \int _0 ^1 B_t ^2 dt\right) = E \exp\left(\frac{\lambda ^2 }{2} \int _0 ^1 (B_1-B_t) ^2 dt\right)= E \exp\left(\lambda \int _0 ^1 B_t dB_t\right) $$ but it did not help me.
Addition
Here are some computations. Set $W_t = C_1 - C _{1-t}$. $W$ is a BM independent on $B$, and
$$ E \exp\left(\frac{\lambda ^2 }{2} \int _0 ^1 (B_t + C_{1-t}) ^2 dt\right) = E \exp\left(\lambda \int _0 ^1( B_t + C_{1-t} )dB_t\right) $$ $$ =E \exp\left(\lambda \int _0 ^1( B_t + W_1 - W_t )dB_t\right) =E \exp\left(\lambda \int _0 ^1 B_t dB_t+ \lambda B_1 W_1 - \lambda \int _0 ^1 W_t dB_t\right) :=F $$
Because $W$ and $B$ are independent, we have $B_1W_1 = \int _0 ^1 W_t dB_t +\int _0 ^1 B_t dW_t$, so
$$ F = E \exp\left(\lambda \int _0 ^1 B_t dB_t + \lambda \int _0 ^1 B_t dW_t\right) = $$ $$ = E \exp\left(\lambda \int _0 ^1 B_t d(B_t + W_t)\right) = E \exp\left(\frac{\lambda ^2 }{2}2\int _0 ^1 B_t ^2 dt\right), $$ the last equality due to relation $\langle B+M,B+M \rangle_t = 2t$. So, it should be
$$ \int _0 ^1 (B_t + C_{1-t})^2 dt \,{\buildrel (d) \over =}\, 2\int _0 ^1 B_t ^2 dt. $$
May it be true?