
Solve $\int_{0}^{\dfrac{\pi}{2}}|sin~x-cos~x\lvert.\text{d}x$
(A) $0$
(B) $2(\surd2-1)$
(C) $\surd2-1$
(D) $2(\surd2+1)$
Answer
162.6k+ views
Hint: In the given question, we can observe that the function given does not have any positive values over its entire range. The value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$. So, the given function is broken into two parts, one from $0$ to $\dfrac{\pi}{4}$ and another from $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
Formula used:
Complete step by step solution: The given integral is: I = $\int_{0}^{\dfrac{\pi}{2}}|sin~x-cos~x\lvert.\text{d}x$
Here the given function is f(x) = $\|sin~x-cos~x\lvert$ and the total range over which the integral is to be done is $0$ to $\dfrac{\pi}{2}$, with a lower limit of $0$ and an upper limit of $\dfrac{\pi}{2}$. We observe that the value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
So, the above integral can be written as I = $\int_{0}^{\dfrac{\pi}{4}}(sin~x-cos~x).\text{d}x+\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}(sin~x-cos~x).\text{d}x$
As integration of $sin x = -cos x$ and the integration of $cos x = sin x$. So, the above integral becomes
I = $\left[-cos~x-sin~x\right]_{0}^{\dfrac{\pi}{4}}+\left[-cos~x-sin~x\right]_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}$
We should remember the following values of sine and cosine functions at different angles: $sin 0 = 0$; sin $\dfrac{\pi}{4}$ = $\dfrac{1}{\surd2}$; sin $\dfrac{\pi}{2}$ = 1; $cos 0 = 1$; cos $\dfrac{\pi}{4}$ =$\dfrac{1}{\surd2}$; cos $\dfrac{\pi}{2}$ = 0.
Using these values to solve the question, we get
I = $(\dfrac{-1}{\surd2}-\dfrac{1}{\surd2}-(-1-0))+(-0-1-(\dfrac{-1}{\surd2}-\dfrac{1}{\surd2})$
I = $2(\surd2-1)$
Hence, $\int_{0}^{\dfrac{\pi}{2}}|sin~x-cos~x\lvert.\text{d}x$ = $2(\surd2-1)$
Thus, Option (B) is correct.
Note: The integration should be done carefully. We should keep in mind that it is a question of definite integration, so we should also substitute the upper and lower limits after integrating.
Formula used:
Complete step by step solution: The given integral is: I = $\int_{0}^{\dfrac{\pi}{2}}|sin~x-cos~x\lvert.\text{d}x$
Here the given function is f(x) = $\|sin~x-cos~x\lvert$ and the total range over which the integral is to be done is $0$ to $\dfrac{\pi}{2}$, with a lower limit of $0$ and an upper limit of $\dfrac{\pi}{2}$. We observe that the value of this function is negative in the range of $0$ to $\dfrac{\pi}{4}$ and positive in the range of $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$.
So, the above integral can be written as I = $\int_{0}^{\dfrac{\pi}{4}}(sin~x-cos~x).\text{d}x+\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}(sin~x-cos~x).\text{d}x$
As integration of $sin x = -cos x$ and the integration of $cos x = sin x$. So, the above integral becomes
I = $\left[-cos~x-sin~x\right]_{0}^{\dfrac{\pi}{4}}+\left[-cos~x-sin~x\right]_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}$
We should remember the following values of sine and cosine functions at different angles: $sin 0 = 0$; sin $\dfrac{\pi}{4}$ = $\dfrac{1}{\surd2}$; sin $\dfrac{\pi}{2}$ = 1; $cos 0 = 1$; cos $\dfrac{\pi}{4}$ =$\dfrac{1}{\surd2}$; cos $\dfrac{\pi}{2}$ = 0.
Using these values to solve the question, we get
I = $(\dfrac{-1}{\surd2}-\dfrac{1}{\surd2}-(-1-0))+(-0-1-(\dfrac{-1}{\surd2}-\dfrac{1}{\surd2})$
I = $2(\surd2-1)$
Hence, $\int_{0}^{\dfrac{\pi}{2}}|sin~x-cos~x\lvert.\text{d}x$ = $2(\surd2-1)$
Thus, Option (B) is correct.
Note: The integration should be done carefully. We should keep in mind that it is a question of definite integration, so we should also substitute the upper and lower limits after integrating.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
