The value of integral $\int_{-2}^0\left(x^3+3 x^2+3 x+5+(x+1) \cos (x+1)\right) d x$ is equal to
The value of integral $\int_{-2}^0\left(x^3+3 x^2+3 x+5+(x+1) \cos (x+1)\right) d x$ is equal to
0
6
4
8
Solution
Let $\begin{aligned} I & =\int_{-2}^0\left[x^3+3 x^2+3 x+5+(x+1) \cos (x+1)\right] d x \\ & =\int_{-2}^0\left[(x+1)^3+4+(x+1) \cos (x+1)\right] d x\end{aligned}$
$\begin{array}{ll}
& \text { Put } x+1=t \Rightarrow d x=d t \\
\therefore \quad & I=\int_{-1}^1\left(t^3+4+t \cos t\right) d t
\end{array}$ Since $t^3$ and $t \cos t$ are odd functions.
$\therefore \quad I=\int_{-1}^1 4 d t=4[t]_{-1}^1=8$
[Note: The answer of the question is not mentioned as an option.]