Question Number 200250 by Calculusboy last updated on 16/Nov/23
Answered by Mathspace last updated on 16/Nov/23
![∫_0 ^x sin(x+t)dt =[cos(x+t)]_(t=0) ^(t=x) =cos(2x)−cosx ⇒(d/dx)(∫_0 ^x sin(x+t)dt) =(d/dx)(cos(2x)−cosx) =−2sin(2x)+sinx](Q200290.png)
$$\int_{\mathrm{0}} ^{{x}} {sin}\left({x}+{t}\right){dt} \\ $$$$=\left[{cos}\left({x}+{t}\right)\right]_{{t}=\mathrm{0}} ^{{t}={x}} ={cos}\left(\mathrm{2}{x}\right)−{cosx} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left(\int_{\mathrm{0}} ^{{x}} {sin}\left({x}+{t}\right){dt}\right) \\ $$$$=\frac{{d}}{{dx}}\left({cos}\left(\mathrm{2}{x}\right)−{cosx}\right) \\ $$$$=−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)+{sinx} \\ $$
Commented by Calculusboy last updated on 16/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Frix last updated on 16/Nov/23
$$\frac{{d}}{{dx}}\underset{\mathrm{0}} {\overset{{x}} {\int}}{f}\left({x},\:{t}\right){dt}=\underset{\mathrm{0}} {\overset{{x}} {\int}}\frac{{d}}{{dx}}{f}\left({x},\:{t}\right){dt}+\underset{{t}\rightarrow{x}} {\mathrm{lim}}\:{f}\left({x},\:{t}\right) \\ $$