Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 10 Maths textbook Solution.

Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given: $\int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx$

Solution:

$\int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx=\int_{0}^{\frac{\pi }{2}}\sin xdx+\int_{0}^{\frac{\pi }{2}}\cos \: xdx$

$=[-\cos x]_{0}^{\frac{\pi}{2}}+[\sin x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \int \sin x d x=-\cos x \\ \int \cos x d x=\sin x \end{array}\right]$

$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\left[\sin \frac{\pi}{2}-\sin 0\right]\left[\begin{array}{l} \cos \frac{\pi}{2}=0, \cos 0=1 \\ \sin \frac{\pi}{2}=1, \sin 0=0 \end{array}\right]$

$=-\left ( 0-1 \right )+\left ( 1-0 \right )$

$=-\left ( -1 \right )+1$

$=1+1=2$

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