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  • Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 15 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 15 Maths Textbook Solution.

Answers (1)

Answer:  4

Hint: We will use the concept of even function to solve the integral.

Given:

                \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\{\sin |x|+\cos |x|\} d x

Solution:

\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\{\sin |x|+\cos |x|\} d x \\ & \end{aligned}

f(-x)=\sin |-x|+\cos |-x|=\sin |x|+\cos |x|=f(x)

                 

\begin{aligned} &f(x) \text { is a even function }\\ & \end{aligned}

I=2 \int_{0}^{\frac{\pi}{2}}(\sin x+\cos x) d x

\begin{aligned} &{\left[\text { for even function } \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x d x=2 \int_{0}^{\frac{\pi}{2}} x d x\right]} \\ & \end{aligned}

I=2[-\cos x+\sin x]_{0}^{\frac{\pi}{2}} \\

I=2[(0+1)-(-1-0)] \\

I=2(2)=4

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