Question Number 47637 by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
![∫_0 ^1 sin([x]+[2x])dx](https://www.tinkutara.com/question/Q47637.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 12/Nov/18
![changement 2x=t give I =(1/2)∫_0 ^2 sin([(t/(2 ))]+[t])dt ⇒ 2I = ∫_0 ^1 sin([(t/2)]+[t])dt +∫_1 ^2 sin([(t/(2 ))]+[t])dt =0 +∫_1 ^2 sin(1)) because [(t/2)]=0 and [t]=1 ⇒I =((sin(1))/2) .](https://www.tinkutara.com/question/Q47661.png)
$${changement}\:\mathrm{2}{x}={t}\:{give}\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {sin}\left(\left[\frac{{t}}{\mathrm{2}\:}\right]+\left[{t}\right]\right){dt}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left(\left[\frac{{t}}{\mathrm{2}}\right]+\left[{t}\right]\right){dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} {sin}\left(\left[\frac{{t}}{\mathrm{2}\:}\right]+\left[{t}\right]\right){dt} \\ $$$$\left.=\mathrm{0}\:+\int_{\mathrm{1}} ^{\mathrm{2}} {sin}\left(\mathrm{1}\right)\right)\:{because}\:\left[\frac{{t}}{\mathrm{2}}\right]=\mathrm{0}\:{and}\:\left[{t}\right]=\mathrm{1}\:\Rightarrow{I}\:=\frac{{sin}\left(\mathrm{1}\right)}{\mathrm{2}}\:. \\ $$
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
![∫_0 ^(1/2) sin([x]+[2x])dx+∫_(1/2) ^1 sin([x]+[2x])dx =∫_0 ^(1/2) sin(0+0)dx+∫_(1/2) ^1 sin(0+1)dx =0+sin1×(1/2) =(1/2)sin1](https://www.tinkutara.com/question/Q47671.png)
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx}+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:{sin}\left(\left[{x}\right]+\left[\mathrm{2}{x}\right]\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\left(\mathrm{0}+\mathrm{0}\right){dx}+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:{sin}\left(\mathrm{0}+\mathrm{1}\right){dx} \\ $$$$=\mathrm{0}+{sin}\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{1} \\ $$