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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.

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

Answers (1)

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

Explanation:

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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