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  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 31

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 31

Answers (1)

Answer:  2\sqrt{2}-2

Hint: To solve this question we will split \sin x.\cos x in two forms.                   \left[\begin{array}{l} \because \sin x=\cos x \\ \tan x=1 \\ x=\frac{\pi}{4} \end{array}\right]

Given:   \int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x

Solution:

\begin{aligned} &\int_{0}^{\frac{\pi}{4}}(\sin x-\cos x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \\ & \end{aligned}

=-[-\cos x-\sin x]_{0}^{\frac{\pi}{4}}+[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}                                                                \left[\begin{array}{l} |x|=-x, x<0 \\ =x, x>0 \end{array}\right]

\begin{aligned} &=-[\cos x-\sin x]_{0}^{\frac{\pi}{4}}-[\cos x+\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ & \end{aligned}

=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(1+0)-\left[(0+1)-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\right] \\

=2 \sqrt{2}-2

 

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