Question

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

Answer 1

We have,

\[I = \int_0^\pi \sin^3 x\left( 1 + 2\cos x \right) \left( 1 + \cos x \right)^2 d x\]

\[ = \int_0^\pi \sin^2 x\left( 1 + 2\cos x \right) \left( 1 + \cos x \right)^2 \sin x d x\]

\[ = \int_0^\pi \left( 1 - \cos^2 x \right)\left( 1 + 2\cos x \right) \left( 1 + \cos x \right)^2 \sin x d x\]

\[\text{Putting }\cos x = t\]

\[ \Rightarrow - \sin x dx = dt\]

\[\text{When }x \to 0; t \to 1\]

\[\text{and }x \to \pi; t \to - 1\]

\[ \therefore I = - \int_1^{- 1} \left( 1 - t^2 \right)\left( 1 + 2t \right) \left( 1 + t \right)^2 dt\]

\[ = \int_{- 1}^1 \left( 1 - t^2 \right)\left( 1 + 2t \right) \left( 1 + t \right)^2 dt\]

\[ = \int_{- 1}^1 \left( 1 + 2t - t^2 - 2 t^3 \right)\left( 1 + 2t + t^2 \right)dt\]

\[ = \int_{- 1}^1 \left( 1 + 2t + t^2 + 2t + 4 t^2 + 2 t^3 - t^2 - 2 t^3 - t^4 - 2 t^3 - 4 t^4 - 2 t^5 \right)dt\]

\[ = \int_{- 1}^1 \left( 1 + 4t + 4 t^2 - 2 t^3 - 5 t^4 - 2 t^5 \right)dt\]

\[ = \left[ t + 2 t^2 + \frac{4 t^3}{3} - \frac{t^4}{2} - t^5 - \frac{t^6}{3} \right]_{- 1}^1 \]

\[ = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} - \left( - 1 \right) - 2 \left( - 1 \right)^2 - \frac{4 \left( - 1 \right)^3}{3} + \frac{\left( - 1 \right)^4}{2} + \left( - 1 \right)^5 + \frac{\left( - 1 \right)^6}{3}\]

\[ = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} + 1 - 2 + \frac{4}{3} + \frac{1}{2} - 1 + \frac{1}{3}\]

\[ = \frac{8}{3}\]