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MHT CET · Maths · Indefinite Integration

\(\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\)
(where \(\mathrm{C}\) is a constant of integration)

  1. A \(-2 \sqrt{1-x}-\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}\)
  2. B \(-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}\)
  3. C \(2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}\)
  4. D \(-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}-\sqrt{x(1-x)}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}\)

Step-by-step Solution

Detailed explanation

\(\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} d x \)
\( \int \frac{1-\sqrt{x}}{\sqrt{1-x}} d x=\int \frac{d x}{\sqrt{1-x}}-\int \frac{\sqrt{x}}{\sqrt{1-x}} d x \)
\( -2 \sqrt{1-x}-\int \frac{\sqrt{\cos ^2 \theta}}{\sqrt{1-\cos ^2 \theta}}(-2 \cos \theta \cdot \sin \theta d \theta)\) \(\left[\operatorname{let} x=\cos ^2 \theta\right] \)
\( =-2 \sqrt{1-x}+\int 2 \cos ^2 \theta d \theta\)
\(=-2 \sqrt{1-x}+\int(1+\cos 2 \theta) d \theta \)
\( =-2 \sqrt{1-x}+\theta+\frac{\sin 2 \theta}{2}+C \)
\( =-2 \sqrt{1-x}+\theta+\sin \theta \cdot \cos \theta+C \)
\( =-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}+C \)
\( =-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C\)