Get Answers to all your Questions

header-bg qa

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 25 Maths Textbook Solution.

Answers (1)

Answer: 2

Hint: Use indefinite formula then put the limit to solve this integral

Given: \int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}


\int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}

=\int_{0}^{\frac{\pi }{2}}\sqrt{2\cos ^{2}\frac{x}{2}dx}                                                                                  \left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]

=\int_{0}^{\frac{\pi }{2}}\sqrt{2}\cos \frac{x}{2}dx                                                                                                    \left [ \int \cos \: axdx=\frac{\sin \: ax}{a} \right ]

\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{2}} \cos \frac{x}{2} d x \\ &=\sqrt{2}\left[\frac{\sin \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}} \\ &=\sqrt{2} .2\left[\sin \frac{x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=2 \sqrt{2}\left[\sin \frac{\pi}{2 \times 2}-\sin \frac{0}{2}\right] \end{aligned}                                                                    \left[\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0\right]

\begin{aligned} &=2 \sqrt{2}\left[\sin \frac{\pi}{4}-0\right] \\ &=2 \sqrt{2}\left[\frac{1}{\sqrt{2}}-0\right] \\ &=2 \sqrt{2} \frac{1}{\sqrt{2}}=2 \end{aligned}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support