Question

\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

Answer 1

\[Let\ I = \int_0^\pi \sin^3 x \left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 d x . Then, \]
\[I = \int_0^\pi \sin x \sin^2 x \left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 d x\]
\[ \Rightarrow I = \int_0^\pi \sin x \left( 1 - \cos^2 x \right)\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 d x\]
\[ \Rightarrow I = \int_0^\pi \sin x \left( 1 - \cos x \right)\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^3 d x\]
\[Let\ \cos x = t . Then, - \sin\ x\ dx\ = dt\]
\[When\ x = 0, t = 1\ and\ x\ = \pi, t = - 1\]
\[ \therefore I = - \int_1^{- 1} \left( 1 - t \right)\left( 1 + 2t \right) \left( 1 + t \right)^3 dt\]
\[ \Rightarrow I = \int_{- 1}^1 \left( 1 + t - 2 t^2 \right)\left( 1 + t^3 + 3t + 3 t^2 \right) dt\]
\[ \Rightarrow I = \int_{- 1}^1 \left( 1 + t^3 + 3t + 3 t^2 + t + t^4 + 3 t^2 + 3 t^3 - 2 t^2 - 2 t^5 - 6 t^3 - 6 t^4 \right) dt\]
\[ \Rightarrow I = \int_{- 1}^1 \left( 1 + 4t + 4 t^2 - 2 t^3 - 5 t^4 - 2 t^5 \right) dt\]
\[ \Rightarrow I = \left[ t + 2 t^2 + \frac{4 t^3}{3} - \frac{t^4}{2} - t^5 - \frac{t^6}{3} \right]_{- 1}^1 \]
\[ \Rightarrow I = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} + 1 - 2 + \frac{4}{3} + \frac{1}{2} - 1 + \frac{1}{3}\]
\[ \Rightarrow I = \frac{8}{3}\]