Question Number 85414 by M±th+et£s last updated on 21/Mar/20
$$\int\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cos}\left({x}\right)}\:}\:{dx} \\ $$
Commented by M±th+et£s last updated on 21/Mar/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cos}\left({x}\right)}}\:{dx}\:\:\:{typo}....\: \\ $$
Answered by mind is power last updated on 23/Mar/20
![∫_0 ^(π/2) (((1−(√(cos(x))))(1+cos(x)))/(1−cos^2 (x)))dx =∫_0 ^(π/2) ((2cos^2 ((x/2)))/(sin^2 (x))).(1−(√(cos(x))))dx =∫_0 ^(π/2) ((2cos^2 ((x/2))(1−(√(cos(x))))dx)/(4sin^2 ((x/2))cos^2 ((x/2)))) =∫_0 ^(π/2) (((1−(√(cos(x)))))/(2sin^2 ((x/2))))dx by part u′=(1/(2sin^2 ((x/2))))=(1/2)(cot^2 ((x/2))+1)⇒u=−cot((x/2)) =[−cot((x/2))(1−(√(cos(x))))]_0 ^(π/2) +∫_0 ^(π/2) ((cot((x/2)).sin(x))/(2(√(cos(x)))))dx =−1+∫_0 ^(π/2) ((cos((x/2))sin(x))/(sin((x/2)).2(√(cos(x))))) =−1+∫_0 ^(π/2) ((cos^2 ((x/2)))/(√(cos(x))))..A A=−1+∫_0 ^(π/2) ((1+cos(x))/(2(√(cos(x))))) =−1+∫_0 ^(π/2) ((((1/2)−sin((x/2))cos((x/2)))/(√(sin(x))))dx =−1+∫_0 ^(π/2) ((1+sin(x))/(2(√(sin(x)))))dx=−1+(1/2){∫_0 ^(π/2) (dx/(√(sin(x))))+∫_0 ^(π/2) (√(sin(x)))dx} ∫_0 ^(π/2) (dx/(√(sin(x)))) ∫_0 ^(π/2) (dx/(√(2sin((x/2))cos((x/2)))))=∫_0 ^(π/2) (dx/(√(1−(sin((x/2))−cos((x/2)))^2 ))) =∫_0 ^(π/2) (dx/(√(1−2sin^2 ((π/4)−(x/2)))))=∫_0 ^(π/4) ((2dr)/(√(1−2sin^2 (r))))=2F((π/4)∣2) ∫_0 ^(π/2) (√(sin(x)))dx=∫_0 ^(π/2) (√(1−(sin((x/2))−cos((x/2)))^2 ))dx =∫_0 ^(π/2) (√(1−2sin^2 ((π/4)−(x/2))))dx =2∫_0 ^(π/4) (√(1−2sin^2 (r)))dr=2E((π/4)∣2) =−1+2(E((π/4)∣2)+F((π/4)∣2))](Q85567.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{1}−\sqrt{{cos}\left({x}\right)}\right)\left(\mathrm{1}+{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{sin}^{\mathrm{2}} \left({x}\right)}.\left(\mathrm{1}−\sqrt{{cos}\left({x}\right)}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt{{cos}\left({x}\right)}\right){dx}}{\mathrm{4}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\: \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{1}−\sqrt{{cos}\left({x}\right)}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:{by}\:{part}\: \\ $$$${u}'=\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left({cot}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\right)\Rightarrow{u}=−{cot}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\left[−{cot}\left(\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt{{cos}\left({x}\right)}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cot}\left(\frac{{x}}{\mathrm{2}}\right).{sin}\left({x}\right)}{\mathrm{2}\sqrt{{cos}\left({x}\right)}}{dx} \\ $$$$=−\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right){sin}\left({x}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right).\mathrm{2}\sqrt{{cos}\left({x}\right)}}\:=−\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\sqrt{{cos}\left({x}\right)}}..{A} \\ $$$${A}=−\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{cos}\left({x}\right)}{\mathrm{2}\sqrt{{cos}\left({x}\right)}} \\ $$$$=−\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\mathrm{1}}{\mathrm{2}}−{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right.}{\sqrt{{sin}\left({x}\right)}}{dx} \\ $$$$=−\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{sin}\left({x}\right)}{\mathrm{2}\sqrt{{sin}\left({x}\right)}}{dx}=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{{sin}\left({x}\right)}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}\right\} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{{sin}\left({x}\right)}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{\mathrm{1}−\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)−{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2}{dr}}{\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({r}\right)}}=\mathrm{2}{F}\left(\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)−{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({r}\right)}{dr}=\mathrm{2}{E}\left(\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right) \\ $$$$=−\mathrm{1}+\mathrm{2}\left({E}\left(\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right)+{F}\left(\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right)\right) \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 23/Mar/20
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$