Explanation

(c)
$I = \int_{- 2}^{0}{\left\lbrack x^{3} + 3x^{2} + 3x + 3 + (x + 1)\cos(x + 1) \right\rbrack dx}$
$= \int_{- 2}^{0}{\left\lbrack (x + 1)^{3} + 2 + (x + 1)\cos(x + 1) \right\rbrack dx}$
Put, $\ x + 1 = t \Rightarrow dx = dt$
$\therefore\ I = \int_{- 1}^{1}{t^{3}dt + 2\int_{- 1}^{1}{dt + \int_{- 1}^{1}{t\cos{t\ dt}}}}$
$= 0 + 2\left\lbrack 1 - ( - 1) \right\rbrack + 0$
$\Rightarrow I = 4\begin{bmatrix}
\because t^{3}and\ t\cos{t\ are\ odd\ functions.} \\
\therefore\ \int_{- 1}^{1}{t^{3}dt = \int_{1}^{1}{t\cos{t\ dt = 0}}}
\end{bmatrix}$