Solveeit Logo
Login

Question

Mathematics Question on Integrals of Some Particular Functions

The value of ∫0aa−xxdx\int^{^a}_{0}\sqrt{\frac{a-x}{x}}dx is

A

a2\frac{a}{2}

B

a4\frac{a}{4}

C

πa2\frac{\pi a}{2}

D

πa4\frac{\pi a}{4}

Answer

πa2\frac{\pi a}{2}

Explanation

Solution

Let x = cos2^2, θ\thetaso that dx = - 2a sin θ\theta cos θ\theta dθ\theta
Now ∫0aa−xxdx=∫π20a−acos2θacos2θ\int^{^a}_{0}\sqrt{\frac{a-x}{x}}dx=\int^{0}_{\frac{\pi}{2}}\sqrt{\frac{a-a cos^{2} \theta}{a cos^{2} \theta}}
(−asinθcosθ)dθ\quad\quad\quad\quad\quad\left(-a sin \theta cos \theta \right)d \theta
[∵θ=π2atx=0;θ=0atx=a]\quad\quad \quad \quad \quad \quad \quad \quad \left[\because \theta=\frac{\pi}{2} at x=0;\theta=0 at x =a \right]
=a∫0π/21−cos2θcos2θ=a\int^{^{\pi/2}}_{_0}\sqrt{\frac{1-cos^{2} \theta}{cos^{2} \theta }} 2 sin θ\theta cos θ\theta d θ\theta
\quad \quad \quad \quad [∵∫t0f(x)dx=−∫t0f(x)dx]\left[\because \int^{0}_{t} f \left(x\right)dx=-\int^{0}_{t} f \left(x\right)dx\right]
=a∫0π/22sinθcosθ.sinθcosθdθ=a\int^{\pi/2}_{0}2\frac{sin \theta }{cos \theta }.sin \theta cos \theta d\theta
=a∫0π/22sin2θdθ=a∫0π/2(1−cos2θ)dθ=a\int^{\pi/2}_{0} 2sin^{2} \theta d\theta =a\int ^{\pi/ 2}_{0} \left(1-cos 2 \theta \right)d \theta
[∵cos2θ=1−2sin2θ]\quad\quad\quad\quad\left[\because cos 2\theta =1-2sin^{2} \theta \right]
=a[θ−sin2θ2]0π/2=a\left[\theta -\frac{sin 2\theta }{2}\right]^{^{\pi/2}}_{_{_0}}
=a[(π2−sinπ2)−(0−0)]=a\left[\left(\frac{\pi}{2}-\frac{sin \pi}{2}\right)-\left(0-0\right)\right]
=a(π2−0)=aπ2=a\left(\frac{\pi}{2}-0\right)=\frac{a\pi}{2}