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$\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x=$
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$\frac{6}{\pi}$
Given $\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x$
Let $I=\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x$
$\begin{aligned} & \text { put } \frac{\pi}{3} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{dx}=\frac{3}{\pi} \mathrm{dt} \\ & \text { and at } \mathrm{x}=0, \mathrm{t}=0 \\ & \text { at } \mathrm{x}=3, \mathrm{t}=\pi \\ & \Rightarrow \mathrm{I}=\int_0^\pi(\sin (\mathrm{t})-\cos (\mathrm{t})) \frac{3}{\pi} \mathrm{dt} \\ & =\frac{3}{\pi}[-\cot (\mathrm{t})-\sin (\mathrm{t})]_0^\pi \\ & =\frac{3}{\pi}[(-(-1)-0)-(-1-0)]=\frac{6}{\pi}\end{aligned}$
Let $I=\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x$
$\begin{aligned} & \text { put } \frac{\pi}{3} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{dx}=\frac{3}{\pi} \mathrm{dt} \\ & \text { and at } \mathrm{x}=0, \mathrm{t}=0 \\ & \text { at } \mathrm{x}=3, \mathrm{t}=\pi \\ & \Rightarrow \mathrm{I}=\int_0^\pi(\sin (\mathrm{t})-\cos (\mathrm{t})) \frac{3}{\pi} \mathrm{dt} \\ & =\frac{3}{\pi}[-\cot (\mathrm{t})-\sin (\mathrm{t})]_0^\pi \\ & =\frac{3}{\pi}[(-(-1)-0)-(-1-0)]=\frac{6}{\pi}\end{aligned}$
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