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Question Number 168966 by cortano1 last updated on 22/Apr/22

$$\underset{x\to 0}{\lim }\frac{{\int }_0^x\left({\int }_0^{u^2}{\tan }^{-1}(1+t)dt\right)dt}{x-x\cos x}=?$$

Answered by bobhans last updated on 22/Apr/22

$$\underset{x\to 0}{\lim }\frac{{\int }_0^x\left({\int }_0^{u^2}{\tan }^{-1}(1+t)dt\right)dt}{x-x\cos x}$$$$=\underset{x\to 0}{\lim }\frac{{\int }_0^{x^2}{\tan }^{-1}(1+t)dt}{1-(\cos x-x\sin x)}=\underset{x\to 0}{\lim }\frac{{\int }_0^{x^2}{\tan }^{-1}(1+t)dt}{1+x\sin x-\cos x}$$$$=\underset{x\to 0}{\lim }\frac{2x{\tan }^{-1}(1+x^2)}{\sin x+x\cos x+\sin x}$$$$=\frac{\pi }{4}.\underset{x\to 0}{\lim }\frac{2x}{2\sin x+x\cos x}=\frac{\pi }{2}\underset{x\to 0}{\lim }\frac{1}{\frac{2\sin x}{x}+\cos x}$$$$=\frac{\pi }{2}.\frac{1}{3}=\frac{\pi }{6}$$

Answered by qaz last updated on 22/Apr/22

$$\arctan (1+\mathrm{t})=\arctan 1+\arctan \frac{\mathrm{t}}{2+\mathrm{t}}=\frac{\pi }{4}+\arctan (1-\frac{1}{1+\frac{\mathrm{t}}{2}})$$$$=\frac{\pi }{4}+\arctan [1-(1-\frac{\mathrm{t}}{2}+...)]=\frac{\pi }{4}+\mathrm{o}\left(1\right)........(\mathrm{t}\to 0)$$$$\Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{{\int }_0^x\left({\int }_0^{u^2}{\tan }^{-1}(1+t)dt\right)dt}{x-x\cos x}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{{\int }_0^{\mathrm{x}}{\int }_0^{{\mathrm{u}}^2}[\frac{\pi }{4}+\mathrm{o}\left(1\right)]\mathrm{dtdu}}{\frac{1}{2}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right)}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{{\int }_0^{\mathrm{x}}[\frac{\pi }{4}{\mathrm{u}}^2+\mathrm{o}\left({\mathrm{u}}^2\right)]\mathrm{du}}{\frac{1}{2}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right)}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{\pi }{12}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right)}{\frac{1}{2}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right)}$$$$=\frac{\pi }{6}$$

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