Get Answers to all your Questions

header-bg qa

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 40 maths textbook solution.

Answers (1)

Answer:-  \frac{1}{\pi}\left(\frac{5}{2}-\frac{1}{\pi}\right)

Hints:-  You must know the integral rules of trignometric functions.

Given:-  \int_{0}^{3 / 2}|x \cos \pi x| d x

Solution : \int_{0}^{3 / 2}|x \cos \pi x| d x

               \begin{aligned} &0<x<\frac{1}{2} \\ &0<\pi x<\frac{\pi}{2} \\ &\cos \pi x>0 \rightarrow x \cos (\pi x)>0 \end{aligned}

               \begin{aligned} &|x \cos \pi x|=x \cos \pi x \\ &\frac{1}{2}<x<\frac{3}{2} \end{aligned}

                 \begin{aligned} &\frac{\pi}{2}<\pi x<\frac{3 \pi}{2} \\ &\cos \pi x<0 \rightarrow x \cos \pi x<0 \\ &|x \cos \pi x|=-x \cos \pi x \end{aligned}

                \begin{aligned} &I=\int_{0}^{3 / 2}|x \cos \pi x| \cdot d x \\ &I=\int_{0}^{3 / 2} x \cos \pi x+\int_{1 / 2}^{3 / 2}-(x \cos \pi x) \\ &I=\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2} x \cos \pi x \cdot d x \end{aligned}

               \begin{aligned} &\int x \cos \pi x=x \frac{\sin \pi x}{\pi}-\int \frac{\sin \pi x}{\pi} \\ &\int x \cos \pi x=\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}} \end{aligned}

              \begin{aligned} &I=\left[\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}}\right]_{0}^{\frac{1}{2}}-\left[\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}}\right]_{\frac{1}{2}}^{3 / 2} \\ &I=\left[\frac{1}{\pi}\left(\frac{1}{2}-0\right)+\frac{1}{\pi^{2}}(0-1)\right]-\left[\frac{1}{\pi}\left(\frac{3}{2}(-1)\right)+\frac{1}{\pi^{2}}(0-0)\right] \end{aligned}

              \begin{aligned} &=\left[\frac{1}{2 \pi}-\frac{1}{\pi^{2}}\right]-\left[\frac{-2}{\pi}\right] \\ &=\left[\frac{5}{2 \pi}-\frac{1}{\pi^{2}}\right] \\ &=\frac{1}{\pi}\left[\frac{5}{2}-\frac{1}{\pi}\right] \end{aligned}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support