$\int_{- 1}^{1} x^{2} \sqrt{4 - x^{2}} d x$ এর মান নির্ণয় কর।

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

ধরি, $x = 2 \sin θ \Rightarrow d x = 2 \cos θ d θ$

x = 1 হলে, $θ = \frac{π}{6}; x = - 1$ হলে, $θ = - \frac{π}{6}$

$\int_{- \frac{π}{6}}^{\frac{π}{6}} 4 \sin^{2} θ \times 2 \cos θ \times 2 \cos θ d θ$

$= \int_{- \frac{π}{6}}^{\frac{π}{6}} 16 \sin^{2} θ \cos^{2} θ d θ = 4 \int_{- \frac{π}{6}}^{\frac{π}{6}} \sin^{2} 2 θ d θ$

$= 2 \int_{- \frac{π}{6}}^{\frac{π}{6}} (1 - \cos 4 θ) d θ = \int [θ - \frac{\sin θ}{4}]_{- \frac{π}{6}}^{\frac{π}{6}}$

$= 2 [\frac{π}{3} - \frac{π}{4} \{\frac{\sqrt{3}}{2} - (- \frac{\sqrt{3}}{2})\}] = \frac{2 π}{3} - \frac{\sqrt{3}}{2}$

1 year ago

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