Put x=2cos2⁡θ+5sin2⁡θ. Then dx=[−4cos⁡θsin⁡θ+10sin⁡θcos⁡θ]dθ=3sin⁡2θdθx=2,5⇒θ=0,π/2∫25 5−xx−2dx=∫0π/2 3cos2⁡θ3sin2⁡θ3sin⁡2θdθ=3∫0π/2 cos⁡θsin⁡θ2sin⁡θcos⁡θdθ=6∫0π/2 cos2⁡θdθ=62∫0π/2 (1+cos⁡2θ)dθ=3θ+sin⁡2θ20π/2=3π2=3π2.

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$\int_{2}^{5} \sqrt{\frac{5 - x}{x - 2}} d x =$

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$π$

$\frac{π}{2}$

$\frac{3 π}{2}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

Put $x = 2 \cos^{2} θ + 5 \sin^{2} θ .$ Then $d x = [- 4 \cos θ \sin θ + 10 \sin θ \cos θ] d θ = 3 \sin 2 θ d θ$

$\begin{array}{l} x = 2, 5 \Rightarrow θ = 0, π / 2 \\ \int_{2}^{5} \sqrt{\frac{5 - x}{x - 2}} d x = \int_{0}^{π / 2} \sqrt{\frac{3 \cos^{2} θ}{3 \sin^{2} θ}} 3 \sin 2 θ d θ = 3 \int_{0}^{π / 2} \frac{\cos θ}{\sin θ} 2 \sin θ \cos θ d θ = 6 \int_{0}^{π / 2} \cos^{2} θ d θ \end{array}$