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$\int_0^a x^4 \sqrt{a^2-x^2} d x=$
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Verified Answer
The correct answer is:
$\frac{\pi}{32} a^6$
Put $x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta$
Now $\int_0^3 x^4 \sqrt{a^2-x^2} d x=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos \theta \cos \theta d \theta$
$=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos ^2 \theta d \theta=a^6 \frac{\Gamma\left(\frac{5}{2}\right) \Gamma\left(\frac{3}{2}\right)}{2 \Gamma 4}=\frac{\pi}{32} a^6$,(Using gamma function).
Now $\int_0^3 x^4 \sqrt{a^2-x^2} d x=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos \theta \cos \theta d \theta$
$=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos ^2 \theta d \theta=a^6 \frac{\Gamma\left(\frac{5}{2}\right) \Gamma\left(\frac{3}{2}\right)}{2 \Gamma 4}=\frac{\pi}{32} a^6$,(Using gamma function).
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