The value ∫36(36−x2)3x4dx is equal to

Let,I=∫36(36−x2)3x4dxSubstituing,x=6sinθ dx=6cosθdθI=∫π6π2(36cos2θ)364sin4θ6cosθdθI=∫π6π2(cos4θsin4θ)dθ=∫π6π2cot2θ(cosec2θ−1)dθ=∫π6π2cot2θcosec2θdθ−∫π6π2cot2dθ=−[cot3θ3]π6π2−∫π6π2(cosec2θ−1)dθ=3+[0−3]+(π2−π6)=π3

The value ∫36(36−x2)3x4dx is equal to

Let,I=∫36(36−x2)3x4dxSubstituing,x=6sinθ dx=6cosθdθI=∫π6π2(36cos2θ)364sin4θ6cosθdθI=∫π6π2(cos4θsin4θ)dθ=∫π6π2cot2θ(cosec2θ−1)dθ=∫π6π2cot2θcosec2θdθ−∫π6π2cot2dθ=−[cot3θ3]π6π2−∫π6π2(cosec2θ−1)dθ=3+[0−3]+(π2−π6)=π3

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The value $\int_{3}^{6} \frac{\sqrt{{(36 - x^{2})}^{3}}}{x^{4}} d x$ is equal to

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$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t, I = \int_{3}^{6} \frac{\sqrt{{(36 - x^{2})}^{3}}}{x^{4}} d x \\ S u b s t i t u i n g, x = 6 \sin θ \\ d x = 6 \cos θ d θ \\ I = \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{{(\sqrt{36 \cos^{2} θ})}^{3}}{6^{4} \sin^{4} θ} 6 \cos θ d θ \\ I = \int_{\frac{π}{6}}^{\frac{π}{2}} (\frac{\cos^{4} θ}{\sin^{4} θ}) d θ \\ = \int_{\frac{π}{6}}^{\frac{π}{2}} \cot^{2} θ (\cos e c^{2} θ - 1) d θ \\ = \int_{\frac{π}{6}}^{\frac{π}{2}} \cot^{2} θ \cos e c^{2} θ d θ - \int_{\frac{π}{6}}^{\frac{π}{2}} \cot^{2} d θ \\ = - {[\frac{\cot^{3} θ}{3}]}_{\frac{π}{6}}^{\frac{π}{2}} - \int_{\frac{π}{6}}^{\frac{π}{2}} (\cos e c^{2} θ - 1) d θ \\ = \sqrt{3} + [0 - \sqrt{3}] + (\frac{π}{2} - \frac{π}{6}) = \frac{π}{3} \end{array}$