Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}5{\left(5-4\cos \mathrm{\theta }\right)}^{1/4}\sin \mathrm{\theta }d\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}5{\left(5-4\cos \theta \right)}^{\frac{1}{4}}\sin \theta d\theta . \\ \mathrm{Let}\left(5-4\cos \theta \right)=t.\mathrm{Then},4\sin \theta d\theta =dt \\ \mathrm{When}\theta =0,t=1\mathrm{and}\theta =\mathrm{\pi },t=9 \\ \therefore I=\frac{5}{4}{\int }_1^9{t}^{\frac{1}{4}}dt \\ \Rightarrow I=\frac{5}{4}{\left[\frac{4t^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}{5}\right]}_1^9 \\ \Rightarrow I=\left(9\sqrt{3}-1\right) \end{gathered}$$