The value of int_0^{frac{pi}{2}}|sin x-cos x| | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$.The function $|\sin x - \cos x|$ changes sign at $x = \frac{\pi}{4}$ in the interval $[0, \fra

Vedclass · Accepted Answer

$2(\sqrt{2}-1)$

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$.The function $|\sin x - \cos x|$ changes sign at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.For $0 \leq x \leq \frac{\pi}{4}$,$\cos x \geq \sin x$,so $|\sin x - \cos x| = \cos x - \sin x$.For $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,$\sin x \geq \cos x$,so $|\sin x - \cos x| = \sin x - \cos x$.Thus,$I = \int_0^{\frac{\pi}{4}}(\cos x - \sin x) d x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x - \cos x) d x$.Evaluating the integrals:$I = [\sin x + \cos x]_0^{\frac{\pi}{4}} + [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$.$I = [(\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)] + [(-\cos \frac{\pi}{2} - \sin \frac{\pi}{2}) - (-\cos \frac{\pi}{4} - \sin \frac{\pi}{4})]$.$I = [(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)] + [(0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})]$.$I = [\frac{2}{\sqrt{2}} - 1] + [-1 + \frac{2}{\sqrt{2}}]$.$I = \sqrt{2} - 1 - 1 + \sqrt{2} = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$.

The value of $\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ is

Similar Questions

The value of $\int_1^4 \log [x] dx$,where $[x]$ is the greatest integer function less than or equal to $x$,is equal to:

Evaluate the definite integral: $\int_0^{\pi^2 / 4} (2 \sin \sqrt{x} + \sqrt{x} \cos \sqrt{x}) \, dx$

$\int_{0}^{4}|x-2| d x=$

The value of $\int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$ is:

The value of the integral $\int_{1}^{3} [x^{2}-2x-2] dx$,where $[x]$ denotes the greatest integer function,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module