Question Number 84713 by john santu last updated on 15/Mar/20
$$\int\overset{\:\:\mathrm{1}} {\:}_{\mathrm{0}} \mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)\:\mathrm{dx}\:? \\ $$
Commented by TANMAY PANACEA last updated on 15/Mar/20
$${is}\:{it}\:\:{cos}^{−\mathrm{1}} \:\:{or}\:{cot}^{−\mathrm{1}} \\ $$
Commented by john santu last updated on 16/Mar/20
$$\mathrm{cos}^{−\mathrm{1}} \:\mathrm{sir} \\ $$
Answered by MJS last updated on 15/Mar/20
![not sure if this is correct... ∫arccos (x^2 −x+1) dx= [t=arccos (x^2 −x+1) → dx=−((sin t)/( (√(4cos t −3))))] =−∫((tsin t)/( (√(4cos t −3))))dt= by parts u′=((sin t)/( (√(4cos t −3)))) → u=−(1/2)(√(4cos t −3)) v=t → v′=1 =(1/2)t(√(4cos t −3))−(1/2)∫(√(4cos t −3)) dt ∫(√(4cos t −3))dt=∫(√(1−8sin^2 (t/2)))dt= and this is an incomplete elliptic integral =2E ((t/2) ∣ 8) ...](https://www.tinkutara.com/question/Q84722.png)
$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{correct}… \\ $$$$\int\mathrm{arccos}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:\rightarrow\:{dx}=−\frac{\mathrm{sin}\:{t}}{\:\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}}\right] \\ $$$$=−\int\frac{{t}\mathrm{sin}\:{t}}{\:\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}}{dt}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\:\:\:\:\:{u}'=\frac{\mathrm{sin}\:{t}}{\:\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}}\:\rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}} \\ $$$$\:\:\:\:\:{v}={t}\:\rightarrow\:{v}'=\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{t}\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}\:{dt} \\ $$$$ \\ $$$$\int\sqrt{\mathrm{4cos}\:{t}\:−\mathrm{3}}{dt}=\int\sqrt{\mathrm{1}−\mathrm{8sin}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}}{dt}= \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{incomplete}\:\mathrm{elliptic}\:\mathrm{integral} \\ $$$$=\mathrm{2E}\:\left(\frac{{t}}{\mathrm{2}}\:\mid\:\mathrm{8}\right) \\ $$$$… \\ $$
Answered by TANMAY PANACEA last updated on 15/Mar/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} {cot}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}\left({x}−\mathrm{1}\right)}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left(\frac{{x}−\left({x}−\mathrm{1}\right)}{\mathrm{1}+{x}\left({x}−\mathrm{1}\right)}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}\:{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left\{\left(\mathrm{1}−{x}\right)−\mathrm{1}\right\}{dx} \\ $$$${using}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}\:{dx} \\ $$$$\int{tan}^{−\mathrm{1}} {xdx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$={xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{I}}=\mathrm{2}\mid{xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${I}=\mathrm{2}\left({tan}^{−\mathrm{1}} \mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\right)=\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−\frac{{ln}\mathrm{2}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−{ln}\mathrm{2} \\ $$