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

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

Answers (1)

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Answer:

\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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