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Question Number 47637 by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx} \\$$

Commented by maxmathsup by imad last updated on 12/Nov/18

$${changement}\:\mathrm{2}{x}={t}\:{give}\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {sin}\left(\left[\frac{{t}}{\mathrm{2}\:}\right]+\left[{t}\right]\right){dt}\:\Rightarrow \\$$$$\mathrm{2}{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left(\left[\frac{{t}}{\mathrm{2}}\right]+\left[{t}\right]\right){dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} {sin}\left(\left[\frac{{t}}{\mathrm{2}\:}\right]+\left[{t}\right]\right){dt} \\$$$$\left.=\mathrm{0}\:+\int_{\mathrm{1}} ^{\mathrm{2}} {sin}\left(\mathrm{1}\right)\right)\:{because}\:\left[\frac{{t}}{\mathrm{2}}\right]=\mathrm{0}\:{and}\:\left[{t}\right]=\mathrm{1}\:\Rightarrow{I}\:=\frac{{sin}\left(\mathrm{1}\right)}{\mathrm{2}}\:. \\$$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

$${thank}\:{you}\:{sir}... \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx}+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:{sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\left(\mathrm{0}+\mathrm{0}\right){dx}+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:{sin}\left(\mathrm{0}+\mathrm{1}\right){dx} \\$$$$=\mathrm{0}+{sin}\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{1} \\$$