Question Number 168966 by cortano1 last updated on 22/Apr/22
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{\:{x}} \:\left(\int_{\mathrm{0}} ^{\:{u}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}\right){dt}\:}{{x}−{x}\:\mathrm{cos}\:{x}}\:=? \\ $$
Answered by bobhans last updated on 22/Apr/22
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{{x}} \left(\int_{\mathrm{0}} ^{{u}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}\right){dt}}{{x}−{x}\mathrm{cos}\:{x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}}{\mathrm{1}−\left(\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}}{\mathrm{1}+{x}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{sin}\:{x}+{x}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x}} \\ $$$$=\:\frac{\pi}{\mathrm{4}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}}{\mathrm{2sin}\:{x}+{x}\mathrm{cos}\:{x}}\:=\:\frac{\pi}{\mathrm{2}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{\frac{\mathrm{2sin}\:{x}}{{x}}\:+\mathrm{cos}\:{x}} \\ $$$$=\:\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\pi}{\mathrm{6}} \\ $$
Answered by qaz last updated on 22/Apr/22
![arctan (1+t)=arctan 1+arctan (t/(2+t))=(π/4)+arctan (1−(1/(1+(t/2)))) =(π/4)+arctan[1−(1−(t/2)+...)]=(π/4)+o(1)........(t→0) ⇒ lim _(x→0) ((∫_0 ^( x) (∫_0 ^( u^2 ) tan^(−1) (1+t)dt)dt )/(x−x cos x)) =lim_(x→0) ((∫_0 ^x ∫_0 ^u^2 [(π/4)+o(1)]dtdu)/((1/2)x^3 +o(x^3 ))) =lim_(x→0) ((∫_0 ^x [(π/4)u^2 +o(u^2 )]du)/((1/2)x^3 +o(x^3 ))) =lim_(x→0) (((π/(12))x^3 +o(x^3 ))/((1/2)x^3 +o(x^3 ))) =(π/6)](https://www.tinkutara.com/question/Q168970.png)
$$\mathrm{arctan}\:\left(\mathrm{1}+\mathrm{t}\right)=\mathrm{arctan}\:\mathrm{1}+\mathrm{arctan}\:\frac{\mathrm{t}}{\mathrm{2}+\mathrm{t}}=\frac{\pi}{\mathrm{4}}+\mathrm{arctan}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}}\right)\: \\ $$$$=\frac{\pi}{\mathrm{4}}+\mathrm{arctan}\left[\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}+…\right)\right]=\frac{\pi}{\mathrm{4}}+\mathrm{o}\left(\mathrm{1}\right)……..\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}\:}\frac{\int_{\mathrm{0}} ^{\:{x}} \:\left(\int_{\mathrm{0}} ^{\:{u}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}\right){dt}\:}{{x}−{x}\:\mathrm{cos}\:{x}}\: \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{\mathrm{x}} \int_{\mathrm{0}} ^{\mathrm{u}^{\mathrm{2}} } \left[\frac{\pi}{\mathrm{4}}+\mathrm{o}\left(\mathrm{1}\right)\right]\mathrm{dtdu}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{\mathrm{x}} \left[\frac{\pi}{\mathrm{4}}\mathrm{u}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{u}^{\mathrm{2}} \right)\right]\mathrm{du}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi}{\mathrm{12}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\frac{\pi}{\mathrm{6}} \\ $$