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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 127 Maths Textbook Solution.

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Answer: -2 \sqrt{1-x}+\cos ^{-1} x+\sqrt{x(1-x)}+C

Given:\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x

Hint: You must know the derivatives and integration of sinx and cosx


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

\begin{aligned} &=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta \\ &=\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \times 2 \sin \theta \cos \theta d \theta \\ &=-2 \int \tan \frac{\theta}{2} \sin \theta \cos \theta d \theta \end{aligned}                                                   \left[\begin{array}{l} \text { put } \sqrt{x}=\cos \theta \\ x=\cos ^{2} \theta \\ d x=2 \cos \theta(-\sin \theta) d \theta \\ d x=-2 \sin \theta \cos \theta d \theta \end{array}\right]

\begin{aligned} &=-2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos \theta d \theta \\ &=-4 \int \sin ^{2} \frac{\theta}{2} \cos \theta d \theta \end{aligned}                                                 \left[\begin{array}{l} \because 1-\cos 2 \theta=2 \sin ^{2} \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]

\begin{aligned} &=-4 \int\left(\frac{1-\cos \theta}{2}\right) \cos \theta d \theta \\ &=-2 \int\left(\cos \theta-\cos ^{2} \theta\right) d \theta \\ &=-2 \int \cos \theta d \theta+2 \int \cos ^{2} \theta d \theta \end{aligned}

\begin{aligned} &=-2 \int \cos \theta d \theta+2 \int \frac{1+\cos 2 \theta}{2} d \theta \\ &=-2 \int \cos \theta d \theta+\int 1 d \theta+\int \cos 2 \theta d \theta \end{aligned}

\begin{aligned} &=-2 \sin \theta+\theta+\frac{\sin 2 \theta}{\theta}+C \\ &=-2 \sin \theta+\theta+\frac{2 \sin \theta \cos \theta}{2}+C \\ &=-2 \sin \theta+\theta+\sin \theta \cos \theta+C \\ &=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}+C \quad\left[\because \cos \theta=\sqrt{1-\sin ^{2} \theta}\right] \\ &=-2 \sqrt{1-x}+\cos ^{-1} x+\sqrt{x(1-x)}+C \end{aligned}

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