0 2 2 + x 2 - x ? d x =

Question

$\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} d x =$

Smart Achievers · Accepted Answer

(1)

Ans. $π + 2$

Explanation :

Put $x = 2 \cos θ$ $\Rightarrow d x = - 2 \sin θ d θ$

At $x = 0$ $\Rightarrow 0 = 2 \cos θ$ $\Rightarrow \cos θ = \cos \frac{π}{2}$ $\Rightarrow θ = \frac{π}{2}$

and $x = 2$ $\Rightarrow 2 = 2 \cos θ$ $\Rightarrow \cos θ = \cos 0 °$ $\Rightarrow θ = 0$

Then, $\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} d x$

$= \int_{π / 2}^{0} \sqrt{\frac{2 + 2 \cos θ}{2 - 2 \cos θ}} (- 2 \sin θ) d θ$

$= - \int_{0}^{π / 2} \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} (- 2 \sin θ) d θ$

$= 2 \int_{0}^{π / 2} \sqrt{\frac{1 + 2 \cos^{2} \frac{θ}{2} - 1}{1 - (1 - 2 \sin^{2} \frac{θ}{2})}} . 2 \sin \frac{θ}{2} \cos \frac{θ}{2} d θ$

$= 2 \int_{0}^{π / 2} \frac{\cos \frac{θ}{2}}{\sin \frac{θ}{2}} 2 \sin \frac{θ}{2} \cos \frac{θ}{2} d θ$

$= 2 \int_{0}^{π / 2} 2 \cos^{2} \frac{θ}{2} d θ$

$= 2 \int_{0}^{π / 2} (1 + \cos θ) d θ$

$= 2 {[θ + \sin θ]}_{0}^{π / 2}$ $= 2 [\frac{π}{2} + 1]$

$= π + 2$

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