The value of the integral ∫αβ 1(x−α)(β−x)dx is

Let tx=αcos2⁡θ+βsin2⁡θ then x−α=(β−α)sin2⁡θ,β−x=(β−α)cos2⁡θand dx=2(β−α)sin⁡θcos⁡θdθAlso x=α⇒θ=0 and , x=β⇒θ=π/2∴ ∫αβ dx(x−α)(β−x)=∫0π/2 2(β−α)sin⁡θcos⁡θ(β−α)sin⁡θcos⁡θdθ=2∫0π/2 dθ=π

The value of the integral ∫αβ 1(x−α)(β−x)dx is

Let tx=αcos2⁡θ+βsin2⁡θ then x−α=(β−α)sin2⁡θ,β−x=(β−α)cos2⁡θand dx=2(β−α)sin⁡θcos⁡θdθAlso x=α⇒θ=0 and , x=β⇒θ=π/2∴ ∫αβ dx(x−α)(β−x)=∫0π/2 2(β−α)sin⁡θcos⁡θ(β−α)sin⁡θcos⁡θdθ=2∫0π/2 dθ=π

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The value of the integral $\int_{α}^{β} \frac{1}{\sqrt{(x - α) (β - x)}} d x$ is

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$π / 2$

$π$

none of these

answer is C.

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Detailed Solution

Let $t x = α \cos^{2} θ + β \sin^{2} θ$ then

$x - α = (β - α) \sin^{2} θ, β - x = (β - α) \cos^{2} θ$

and $d x = 2 (β - α) \sin θ \cos θ d θ$

Also $x = α \Rightarrow θ = 0$ and $, x = β \Rightarrow θ = π / 2$

$\begin{array}{l} ∴ \int_{α}^{β} \frac{d x}{\sqrt{(x - α) (β - x)}} \\ = \int_{0}^{π / 2} \frac{2 (β - α) \sin θ \cos θ}{(β - α) \sin θ \cos θ} d θ = 2 \int_{0}^{π / 2} d θ = π \end{array}$