Integration Questions

Question Number 37036 by ajfour last updated on 08/Jun/18

$$\int_{\mathrm{0}} ^{\:\:{a}} \left(\mathrm{1}−\frac{{b}−{x}}{\sqrt{\left({b}−{x}\right)^{\mathrm{2}} +{cx}}}\right)\:{dx}\:=\:? \\$$

Commented by math khazana by abdo last updated on 09/Jun/18

$${I}\:={a}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{a}} \:\:\:\frac{\mathrm{2}\left({b}−{x}\right)\:+\mathrm{2}{c}}{\sqrt{\left({b}−{x}\right)^{\mathrm{2}} \:+{c}}}{dx} \\$$$$={a}\:−\int_{\mathrm{0}} ^{{a}} \:\:\:\:\:\:\frac{\mathrm{2}\left({b}−{x}\right)\:+{c}}{\mathrm{2}\sqrt{\left({b}−{x}\right)^{\mathrm{2}} \:+{cx}}}{dx}\:−\frac{{c}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{a}} \:\:\:\:\:\frac{{dx}}{\sqrt{\left({b}−{x}\right)^{\mathrm{2}} \:+{cx}}} \\$$$$={a}−\sqrt{\left({b}−{x}\right)^{\mathrm{2}} \:+{cx}}\:\:−\frac{{c}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{a}} \:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:−\mathrm{2}{bx}\:+{b}^{\mathrm{2}} \:+{cx}}} \\$$$${but}\:{x}^{\mathrm{2}} \:−\mathrm{2}{bx}\:+{b}^{\mathrm{2}} \:+{cx}\:=\:{x}^{\mathrm{2}} \:+\left({c}−\mathrm{2}{b}\right){x}\:+{b}^{\mathrm{2}} \\$$$$=\:{x}^{\mathrm{2}} \:\:+\mathrm{2}\:\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\:{x}\:+\:\frac{\left({c}−\mathrm{2}{b}\right)^{\mathrm{2}} }{\mathrm{4}}\:+{b}^{\mathrm{2}} \:−\frac{\left({c}−\mathrm{2}{b}\right)^{\mathrm{2}} }{\mathrm{4}} \\$$$$=\left({x}\:+\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{4}{b}^{\mathrm{2}} \:−{c}^{\mathrm{2}} \:+\mathrm{4}{bc}\:−\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{4}} \\$$$$=\left({x}\:+\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:+\:\frac{\mathrm{4}{bc}\:−{c}^{\mathrm{2}} }{\mathrm{4}} \\$$$${case}\:\mathrm{1}\:\:{if}\:\mathrm{4}{bc}\:−{c}^{\mathrm{2}} >\mathrm{0}\:{we}\:{use}\:{the}\:{changement} \\$$$${x}+\:\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}{\mathrm{2}}\:{t}\:\:\:{and} \\$$$$\int_{\mathrm{0}} ^{{a}} \:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\left({c}−\mathrm{2}{b}\right){x}\:+{b}^{\mathrm{2}} }}\:=\:\int_{\frac{{c}−\mathrm{2}{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}} ^{\frac{\mathrm{2}{a}+{c}−{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}} \:\:\:\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}{\mathrm{2}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}{\mathrm{2}}{dt} \\$$$$=\:\left[{argsht}\right]_{\frac{{c}−\mathrm{2}{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}} ^{\frac{\mathrm{2}{a}+{c}−{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} \:}}} \:\:\:\:={ln}\left(\:\frac{\mathrm{2}{a}+{c}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{a}+{c}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}\right)^{\mathrm{2}} }\:\right) \\$$$$−{ln}\left(\:\frac{{c}−\mathrm{2}{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}\:+\sqrt{\mathrm{1}+\left(\frac{{c}−\mathrm{2}{b}}{\sqrt{\mathrm{4}{bc}−{c}^{\mathrm{2}} }}\right)^{\mathrm{2}} }\right)\:={A}\:{so} \\$$$${I}\:=\:{a}−\sqrt{\left({b}−{x}\right)^{\mathrm{2}} \:+{cx}}\:\:−\frac{{c}}{\mathrm{2}}\:{A} \\$$$${case}\mathrm{2}\:\mathrm{4}{bc}−{c}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\left({x}\:+\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}{bc}−{c}^{\mathrm{2}} }{\mathrm{4}} \\$$$$=\left({x}\:+\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\:\left(\sqrt{\frac{{c}^{\mathrm{2}} −\mathrm{4}{bc}}{\mathrm{4}}}\right)^{\mathrm{2}} \:{and}\:{we}\:{use} \\$$$${the}\:{chang}.\:{x}+\frac{{c}−\mathrm{2}{b}}{\mathrm{2}}\:=\frac{\sqrt{{c}^{\mathrm{2}} \:−\mathrm{4}{bc}}}{\mathrm{2}}\:\:\:{u}\:\:...{be}\:{continued}... \\$$

Commented by ajfour last updated on 09/Jun/18

$${thank}\:{you}\:{sir},\:{am}\:{trying}\:{to} \\$$$${comprehend}.. \\$$

Commented by math khazana by abdo last updated on 09/Jun/18

$${you}\:{are}\:{welcome}\:{sir}\:{Ajfour} \\$$