First slide

Methods of integration

Question

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} dx$ is equal to

Moderate

A

$- 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x + x^{2}} + C$
B

$2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x - x^{2}} + C$
C

$- 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x - x^{2}} + C$
D

None of the above

Solution

$Let t = \cos^{- 1} \sqrt{x}$
$\begin{array}{l} Now, \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} dx = \int \sqrt{\frac{1 - \cos t}{1 + \cos t}} \cdot [2 \cos t (- \sin t)] dt \\ \Rightarrow \begin{array}{l} I = - 2 \int \sqrt{\frac{2 \sin^{2} t / 2}{2 \cos^{2} t / 2}} \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2} \cdot \cos tdt \\ [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}, 1 + \cos x = 2 \cos^{2} \frac{x}{2} and \sin 2 t = 2 \sin tcos t \end{array} \end{array}$
$\begin{array}{l} = - 4 \int \sin^{2} \frac{t}{2} \cdot \cos tdt = - 4 \int \frac{1 - \cos t}{2} \cdot \cos tdt \\ = - 2 \int (\cos t - \cos^{2} t) dt \\ = - 2 \int (\cos t - \frac{1 + \cos 2 t}{2}) dt \\ = - 2 \sin t + t + \frac{1}{2} \sin 2 t + C \end{array}$
$\begin{array}{l} = - 2 \sin t + t + \frac{1}{2} \times 2 \sin tcos t + C \\ = - 2 \sqrt{1 - \cos^{2} t} + t + \frac{1}{2} [2 \cdot \sqrt{1 - \cos^{2} t} \cdot \cos t] + C \\ = - 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x} \cdot \sqrt{1 - x} + C \\ = - 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x - x^{2}} + C \end{array}$
$∴ \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} dx = - 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x - x^{2}} + C$

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