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Question Number 169743 by naka3546 last updated on 07/May/22

$$\underset{x\to 0}{\lim }\frac{\underset{0}{\int }\stackrel{x}{}\sqrt{1+\sin t}dt}{x}=a,$$$$then{a}^2-1=...?$$

Answered by Mathspace last updated on 07/May/22

$$byhosital{lim}_{x\to 0}\frac{{\int }_0^x\sqrt{1+sint}dt}{x}$$$$={lim}_{x\to 0}\frac{\sqrt{1+sinx}}{1}=1\Rightarrow a=1\Rightarrow$$$$a^2-1=0$$

Commented by naka3546 last updated on 07/May/22

$$\mathrm{Thank}\mathrm{you},sir.$$

Answered by cortano1 last updated on 08/May/22

$$2^{nd}way$$$$L=\underset{x\to 0}{\lim }\frac{{\int }_0^x\sqrt{1+\sin t}dt}{x}$$$$=\underset{x\to 0}{\lim }\frac{{\int }_0^x(\sin \frac{1}{2}t+\cos \frac{1}{2}t)dt}{x}$$$$=\underset{x\to 0}{\lim }\frac{-2\cos \frac{1}{2}x+2\sin \frac{1}{2}x+2}{x}$$$$=\underset{x\to 0}{\lim }\frac{2(\sin \frac{1}{2}x+2\sin {}^2\frac{1}{4}x)}{x}$$$$=2(\frac{1}{2}+2.\frac{1}{4}.0)=1=a$$

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