$\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ has the value

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Aicharya · Accepted Answer

$\begin{aligned} I & =\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x \ & =\int_0^{\frac{\pi}{4}}-(\sin x-\cos x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x\end{aligned}$ $\cdots\left\{\begin{aligned} \mathrm{f}(x) & =\mathrm{f}(-x) & & 0 \leq x \leq \frac{\pi}{4} \ & =\mathrm{f}(x) & & \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\end{aligned}ight\}$ $\begin{aligned} & =\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \ & =(\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ & =\left[\sin \frac{\pi}{4}+\cos \frac{\pi}{4}-\sin 0-\cos 0ight]\end{aligned}$ $-\left[\cos \frac{\pi}{2}+\sin \frac{\pi}{2}-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}ight]$ $\begin{aligned} & =\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1ight]-\left[0+1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}ight] \ & =\left[\frac{2}{\sqrt{2}}-1ight]-\left[1-\frac{2}{\sqrt{2}}ight] \ & =\frac{4}{\sqrt{2}}-2 \ & =\frac{4 \sqrt{2}}{2}-2 \ & =2 \sqrt{2}-2 \ & =2(\sqrt{2}-1)\end{aligned}$

$\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ has the value

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