#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 26 Maths Textbook Solution.

$\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C$

Given:

$\int \sqrt{\frac{x}{1-x}} d x$ is equal to:

Hint:

You must know about the $\int \sin ^{2} \theta d \theta \text { . }$.

Explanation:

Let I      $=\int \sqrt{\frac{x}{1-x}}$

$=\int \sqrt{\frac{\sin ^{2} \theta}{1-\sin ^{2} \theta} \times} 2 \sin \theta \cos \theta d \theta$                                        $\text { [ Put } \left.x=\sin ^{2} \theta \Rightarrow d x=2 \sin \theta \cos \theta d \theta\right]$

$=\int \sqrt{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}} \cdot 2 \sin \theta \cos \theta d \theta$                                                 $\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

\begin{aligned} &=\int \sqrt{\tan ^{2} \theta} \cdot 2 \sin \theta \cos \theta d \theta \\ &=\int \frac{\sin \theta}{\cos \theta} \times 2 \sin \theta \cos \theta d \theta \\ &=\int 2 \sin ^{2} \theta d \theta \\ &=\int(1-\cos 2 \theta) d \theta \end{aligned}                                            $\left[\because 2 \sin ^{2} \theta=1-\cos 2 \theta\right]$

\begin{aligned} &=\int 1 d \theta-\int \cos 2 \theta d \theta \\ &=\theta-\frac{1}{2} \sin 2 \theta+C \\ &=\theta-\frac{1}{2} \times 2 \sin \theta \cos \theta+C \\ &=\theta-\sin \theta \cos \theta+C \\ &=\theta-\sin \theta \sqrt{1-\sin ^{2} \theta}+C \\ &=\sin ^{-1} \sqrt{x}-\sqrt{x} \sqrt{1-x}+C \\ &=\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C \end{aligned}

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