Question Number 42801 by maxmathsup by imad last updated on 02/Sep/18
$${find}\:{f}\left({x}\right)\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{cosxdx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\:+\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 04/Sep/18
![let I = ∫_(π/4) ^(π/3) ((cosxdx)/(2(1−sin^2 x) +sin^2 x +1)) changement sinx =t give I = ∫_(1/( (√2))) ^((√3)/2) (dt/(2(1−t^2 )+t^2 +1)) = ∫_(1/( (√2))) ^((√3)/2) (dt/(3 −t^2 )) =∫_(1/( (√2))) ^((√3)/2) (dt/(((√3)−t)((√3)+t)))dt = (1/(2(√3))) ∫_(1/( (√2))) ^((√3)/2) {(1/( (√3)−t)) +(1/( (√3)+t))}dt =(1/(2(√3))) [ln∣(((√3)+t)/( (√3)−t))∣]_(1/( (√2))) ^((√3)/2) = (1/(2(√3))){ ln∣(((√3)+((√3)/2))/( (√3)−((√3)/2)))∣ −ln∣(((√3)+(1/( (√2))))/( (√3)−(1/( (√2)))))∣} = (1/(2(√3))){ ln(((3(√3))/( (√3))))−ln((((√6)+1)/( (√6)−1)))} =(1/(2(√3))){ln(3)−ln(((1+(√6))/(−1+(√6))))} .](https://www.tinkutara.com/question/Q42915.png)
$${let}\:{I}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{cosxdx}}{\mathrm{2}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\:+{sin}^{\mathrm{2}} {x}\:+\mathrm{1}}\:{changement}\:{sinx}\:={t}\:{give} \\ $$$${I}\:=\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{3}\:−{t}^{\mathrm{2}} }\:=\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\left(\sqrt{\mathrm{3}}−{t}\right)\left(\sqrt{\mathrm{3}}+{t}\right)}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−{t}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+{t}}\right\}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\left[{ln}\mid\frac{\sqrt{\mathrm{3}}+{t}}{\:\sqrt{\mathrm{3}}−{t}}\mid\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{\:{ln}\mid\frac{\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\mid\right. \\ $$$$\left.−{ln}\mid\frac{\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\mid\right\}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{\:{ln}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\right)−{ln}\left(\frac{\sqrt{\mathrm{6}}+\mathrm{1}}{\:\sqrt{\mathrm{6}}−\mathrm{1}}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{{ln}\left(\mathrm{3}\right)−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{6}}}{−\mathrm{1}+\sqrt{\mathrm{6}}}\right)\right\}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
![∫_(1/( (√2))) ^((√3)/2) (dt/(2(1−t^2 )+t^2 +1)) t=sinx dt=cosxdx ∫_(1/( (√2))) ^((√3)/2) (dt/(3−t^2 ))=∣(1/(2(√3)))ln(((t+(√3))/(t−(√3))))∣_(1/( (√2))) ^((√3)/2) (1/(2(√3))){ln∣(((((√3)/2)+(√3))/(((√3)/2)−(√3))))∣−ln∣((((1/( (√2)))+(√3))/((1/( (√2)))−(√3))))∣} =(1/(2(√3))){ln∣(((3/2)/(−(1/2))))∣−ln∣(((1+(√6))/(1−(√6))))∣} =(1/(2(√3))){ln∣−3∣−ln∣(((1+(√6))/(1−(√6))))} use formula ∫(dx/(a^2 −x^2 ))=(1/(2a))∫((a+x+a−x)/((a+x)(a−x)))dx (1/(2a))[∫(dx/(a−x))+∫(dx/(a+x))] (1/(2a))[∫(dx/(x+a))−∫(dx/(x−a))]=(1/(2a))ln(((x+a)/(x−a)))](https://www.tinkutara.com/question/Q42846.png)
$$\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{t}^{\mathrm{2}} +\mathrm{1}}\:\:{t}={sinx}\:\:{dt}={cosxdx} \\ $$$$\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{3}−{t}^{\mathrm{2}} }=\mid\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\left(\frac{{t}+\sqrt{\mathrm{3}}}{{t}−\sqrt{\mathrm{3}}}\right)\mid_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{{ln}\mid\left(\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\sqrt{\mathrm{3}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\sqrt{\mathrm{3}}}\right)\mid−{ln}\mid\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{3}}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\sqrt{\mathrm{3}}}\right)\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{{ln}\mid\left(\frac{\frac{\mathrm{3}}{\mathrm{2}}}{−\frac{\mathrm{1}}{\mathrm{2}}}\right)\mid−{ln}\mid\left(\frac{\mathrm{1}+\sqrt{\mathrm{6}}}{\mathrm{1}−\sqrt{\mathrm{6}}}\right)\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{{ln}\mid−\mathrm{3}\mid−{ln}\mid\left(\frac{\mathrm{1}+\sqrt{\mathrm{6}}}{\mathrm{1}−\sqrt{\mathrm{6}}}\right)\right\} \\ $$$$ \\ $$$${use}\:{formula}\:\int\frac{{dx}}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{a}}\int\frac{{a}+{x}+{a}−{x}}{\left({a}+{x}\right)\left({a}−{x}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}\left[\int\frac{{dx}}{{a}−{x}}+\int\frac{{dx}}{{a}+{x}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}\left[\int\frac{{dx}}{{x}+{a}}−\int\frac{{dx}}{{x}−{a}}\right]=\frac{\mathrm{1}}{\mathrm{2}{a}}{ln}\left(\frac{{x}+{a}}{{x}−{a}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$