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Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 24 Maths Textbook Solution.

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Answer: 2

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

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


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

=\int_{0}^{\frac{\pi}{2}}\left(\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) d x \quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]

=\int_{0}^{\frac{\pi}{2}} \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x \; \; \; \; \; \;\; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right]

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ &=\int_{0}^{\frac{\pi}{3}} \sin \frac{x}{2} d x+\int_{0}^{\frac{\pi}{3}} \cos \frac{x}{2} d x \end{aligned}                                                    \left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]

\begin{aligned} &=\left[\frac{-\cos \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{\sin \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}} \\ &=-2\left[\cos \frac{x}{2}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin \frac{x}{2}\right]_{0}^{\frac{\pi}{2}} \end{aligned}

\begin{aligned} &=-2\left[\cos \frac{\pi}{2 \times 2}-\cos 0 \times \frac{1}{2}\right]+2\left[\sin \frac{\pi}{2 \times 2}-\sin 0 \times \frac{1}{2}\right\rfloor \\ &=-2\left[\cos \frac{\pi}{4}-\cos 0\right]+2\left[\sin \frac{\pi}{4}-\sin 0\right] \\ &=-2\left[\frac{1}{\sqrt{2}}-1\right]+\left[\frac{1}{\sqrt{2}}-0\right] \end{aligned}                            \left[\begin{array}{l} \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ \sin 0=0 \end{array}\right]


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