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प्रश्न
उत्तर
\[Let\ I = \int_0^\pi 5 \left( 5 - 4\cos \theta \right)^\frac{1}{4} \sin \theta\ d \theta . \]
\[Let\left( 5 - 4 \cos \theta \right) = t . Then, 4 \sin \theta\ d\theta = dt\]
\[When\ \theta = 0, t = 1\ and\ \theta = \pi, t = 9\]
\[ \therefore I = \frac{5}{4} \int_1^9 t^\frac{1}{4} dt\]
\[ \Rightarrow I = \frac{5}{4} \left[ \frac{4 t^\frac{5}{4}}{5} \right]_1^9 \]
\[ \Rightarrow I = \left( 9\sqrt{3} - 1 \right)\]
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