Find \[\int_0^\pi {\left| {\cos x} \right|} \,dx = \]
A. \[\dfrac{1}{2}\]
B. \[ - 2\]
C. \[1\]
D. \[ - 1\]
E. \[2\]
Answer
297.3k+ views
Hint: We have been given to calculate the definite integral of a trigonometric function, which is a cosine function. As we see that the given function is a modulus function, we first break the integral into two parts, then integrate the given function, and finally apply upper and lower limits of integration.
Formula used:
We have been using the following formulas:
1. \[\int {\cos x = \sin x + c} \]
Complete step-by-step solution:
We are given that \[\int_0^\pi {\left| {\cos x} \right|} \,dx\]
Now we divide the integral into two parts, and we get
\[\int_0^\pi {\left| {\cos x} \right|} \,dx = \int_0^{\dfrac{\pi }{2}} {\left| {\cos x} \right|} \,dx + \int_{\dfrac{\pi }{2}}^\pi {\left| {\cos x} \right|} \,dx\]
Now we know that the function
\[\left|cos x \right| = cos x\] if \[ 0 < x < \pi/2\]
\[\left|cos x \right| = - cos x\] if \[\pi/2 < x < \pi \]
Therefore, we get
\[\int_0^\pi {\left| {\cos x} \right|} \,dx = \int\limits_0^{\dfrac{\pi }{2}} {\cos x\,dx - } \int\limits_{\dfrac{\pi }{2}}^\pi {\cos x\,dx...\left( 1 \right)} \]
Now we know that \[\int {\cos x = \sin x} \]
Therefore, by integrating equation (1), we get
\[
\int_0^\pi {\left| {\cos x} \right|} \,dx = \left[ {\sin x} \right]_0^{\dfrac{\pi }{2}} - \left[ {\sin x} \right]_{\dfrac{\pi }{2}}^\pi \\
= \left[ {\sin \dfrac{\pi }{2} - \sin \,0} \right] - \left[ {\sin \pi - \sin \dfrac{\pi }{2}} \right] \\
\]
Now we know that
\[
\sin \dfrac{\pi }{2} = 1 \\
\sin \pi = 0 \\
\sin 0 = 0
\]
Now by substituting these values and simplifying it, we get
\[
\int_0^\pi {\left| {\cos x} \right|} \,dx = \left[ {1 - 0} \right] - \left[ {0 - 1} \right] \\
= 1 - 0 - 0 + 1 \\
= 1 + 1 \\
= 2
\]
Therefore, the value of \[\int_0^\pi {\left| {\cos x} \right|} \,dx\] is equal to \[2\]
Hence, option (E) is correct option
Additional information: The area bound under the function graph \[y = f\left( x \right)\], and above the x-axis, which is bound between two bounds as \[x = a\] and \[x = b\], is the definite integral of a function \[f\left( x \right)\]. In this case, \[a = 0\] and \[b = \dfrac{\pi }{2}\]. One of the capabilities of integral calculus is that when the function is integrated with respect to \[dx\], one of the two boundaries of the area to be found is a curve and the other is the x-axis.
Note: In such questions, the key concept is to always remember all of the properties of definite integrals then simplify the integral, and also remember the integration formula of trigonometric function and values then simplify for the required result.
Formula used:
We have been using the following formulas:
1. \[\int {\cos x = \sin x + c} \]
Complete step-by-step solution:
We are given that \[\int_0^\pi {\left| {\cos x} \right|} \,dx\]
Now we divide the integral into two parts, and we get
\[\int_0^\pi {\left| {\cos x} \right|} \,dx = \int_0^{\dfrac{\pi }{2}} {\left| {\cos x} \right|} \,dx + \int_{\dfrac{\pi }{2}}^\pi {\left| {\cos x} \right|} \,dx\]
Now we know that the function
\[\left|cos x \right| = cos x\] if \[ 0 < x < \pi/2\]
\[\left|cos x \right| = - cos x\] if \[\pi/2 < x < \pi \]
Therefore, we get
\[\int_0^\pi {\left| {\cos x} \right|} \,dx = \int\limits_0^{\dfrac{\pi }{2}} {\cos x\,dx - } \int\limits_{\dfrac{\pi }{2}}^\pi {\cos x\,dx...\left( 1 \right)} \]
Now we know that \[\int {\cos x = \sin x} \]
Therefore, by integrating equation (1), we get
\[
\int_0^\pi {\left| {\cos x} \right|} \,dx = \left[ {\sin x} \right]_0^{\dfrac{\pi }{2}} - \left[ {\sin x} \right]_{\dfrac{\pi }{2}}^\pi \\
= \left[ {\sin \dfrac{\pi }{2} - \sin \,0} \right] - \left[ {\sin \pi - \sin \dfrac{\pi }{2}} \right] \\
\]
Now we know that
\[
\sin \dfrac{\pi }{2} = 1 \\
\sin \pi = 0 \\
\sin 0 = 0
\]
Now by substituting these values and simplifying it, we get
\[
\int_0^\pi {\left| {\cos x} \right|} \,dx = \left[ {1 - 0} \right] - \left[ {0 - 1} \right] \\
= 1 - 0 - 0 + 1 \\
= 1 + 1 \\
= 2
\]
Therefore, the value of \[\int_0^\pi {\left| {\cos x} \right|} \,dx\] is equal to \[2\]
Hence, option (E) is correct option
Additional information: The area bound under the function graph \[y = f\left( x \right)\], and above the x-axis, which is bound between two bounds as \[x = a\] and \[x = b\], is the definite integral of a function \[f\left( x \right)\]. In this case, \[a = 0\] and \[b = \dfrac{\pi }{2}\]. One of the capabilities of integral calculus is that when the function is integrated with respect to \[dx\], one of the two boundaries of the area to be found is a curve and the other is the x-axis.
Note: In such questions, the key concept is to always remember all of the properties of definite integrals then simplify the integral, and also remember the integration formula of trigonometric function and values then simplify for the required result.
Recently Updated Pages
Geometry of Complex Numbers Explained

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Hydrogen and Its Type Important Concepts and Tips for JEE Exam Preparation

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Geostationary and Geosynchronous Satellites

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

