Question

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

Answer 1

\[Let I = \int_0^\pi x \sin x \cos^4 x d x ................(1)\]
\[ = \int_0^\pi \left( \pi - x \right) \sin\left( \pi - x \right) \cos^4 \left( \pi - x \right) d x\]
\[ = \int_0^\pi \left( \pi - x \right) \sin x \cos^4 x dx ..................(2) \]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\pi \left( x + \pi - x \right) \sin x \cos^4 x\ dx \]
\[ = \pi \int_0^\pi \sin x \cos^4 x\ d x \]
\[ Let\ \cos x = t, \text{Then }- sinx \ dx = dt, \]
\[ \text{When} x = 0, t = 1, x = \pi, t = - 1\]
\[\text{Therefore}, 2I = - \pi \int_1^{- 1} t^4 dt\]
\[ = \pi \int_{- 1}^1 t^4 dt\]
\[ = \pi \left[ \frac{t^5}{5} \right]_{- 1}^1 \]
\[ = \frac{\pi}{5} + \frac{\pi}{5}\]
\[ = \frac{2\pi}{5}\]
\[\text{Hence } I = \frac{\pi}{5}\]