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Question: Answered & Verified by Expert
$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$
MathematicsDefinite IntegrationTS EAMCETTS EAMCET 2020 (09 Sep Shift 2)
Options:
  • A $\frac{5^9}{2} \frac{\pi}{2}$
  • B $\frac{16\left(5^9\right)}{63}$
  • C $\frac{3\left(5^{10}\right)}{256} \pi$
  • D $\frac{16\left(5^{10}\right)}{693}$
Solution:
1077 Upvotes Verified Answer
The correct answer is: $\frac{3\left(5^{10}\right)}{256} \pi$
Let $\begin{aligned} l & =\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x \\ & =2 \int_0^5 x^4\left(25-x^2\right)^{5 / 2} d x\end{aligned}$
Put, $x=5 \sin \theta \Rightarrow d x=5 \cos \theta d \theta$
$\therefore \quad I=2 \int_0^{\pi / 2} 5^4 \sin ^4 \theta 5^5 \cdot \cos ^5 \theta \cdot 5 \cos \theta d \theta$
$\begin{aligned} & =2 \times 5^{10} \int_0^{\pi / 2} \sin ^4 \theta \cos ^6 \theta d \theta \\ & =2 \times 5^{10} \int_0^{\pi / 2}\left(1-\cos ^2 \theta\right)^2 \cos ^6 \theta d \theta \\ & =2 \times 5^{10} \int_0^{\pi / 2}\left(\cos ^6 \theta-2 \cos ^8 \theta+\cos ^{10} \theta\right) d \theta \\ & =2 \times 5^{10}\left[\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2}-2 \times \frac{7}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2}\right. \\ & \left.=2 \times \frac{9}{10} \times \frac{7}{8} \times \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2}\right] \frac{\pi}{2} \\ & =2 \times \frac{5^{10} \times \pi \times 5 \times 3}{6 \times 4 \times 2 \times 2} \times \frac{80-140+63}{6 \times 4 \times 2}\left[1-\frac{7}{4}+\frac{63}{80}\right] \\ & =2 \times \frac{5^{10} \times \pi \times 5 \times 3}{6 \times 4 \times 2 \times 2} \times \frac{3}{80}=\frac{3\left(5^{10}\right) \pi}{256}\end{aligned}$

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