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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 24

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 24

Answers (1)

Answer:  6

Hint: We will check for the nature of function ( even or odd) then will use the property of definition

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

Given:  \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x

Solution:

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

                \mathrm{f}(\mathrm{x})=2 \sin |\mathrm{x}|+\cos |\mathrm{x}| \\

                \begin{aligned} \mathrm{f}(-\mathrm{x})=2 \sin |-\mathrm{x}|+\cos |-\mathrm{x}| \\ \end{aligned}

                            =2 \sin |\mathrm{x}|+\cos |\mathrm{x}|=\mathrm{f}(\mathrm{x})

This shows that f(x) is an even function.

So we use the property :

\begin{aligned} &\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ & \end{aligned}

\text { if } f(-x)=f(x)

\begin{aligned} &I=2 \int_{0}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x \\ \end{aligned}

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

    \begin{aligned} &=-2(\cos x \times 2)_{0}^{\frac{\pi}{2}}+2(\sin x)_{0}^{\frac{\pi}{2}} \\ & \end{aligned}

    =-4(0-1)+2(1-0)=(4+2)=6

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