Get Answers to all your Questions

header-bg qa

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


                \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}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support