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Question Number 200250 by Calculusboy last updated on 16/Nov/23

Answered by Mathspace last updated on 16/Nov/23

$${\int }_0^{x}sin(x+t)dt$$$$={\left[cos(x+t)\right]}_{t=0}^{t=x}=cos\left(2x\right)-cosx$$$$\Rightarrow \frac{d}{dx}\left({\int }_0^{x}sin(x+t)dt\right)$$$$=\frac{d}{dx}(cos\left(2x\right)-cosx)$$$$=-2sin\left(2x\right)+sinx$$

Commented by Calculusboy last updated on 16/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Answered by Frix last updated on 16/Nov/23

$$\frac{d}{dx}\underset{0}{\stackrel{x}{\int }}f(x,t)dt=\underset{0}{\stackrel{x}{\int }}\frac{d}{dx}f(x,t)dt+\underset{t\to x}{\lim }f(x,t)$$

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