Question Number 186637 by Mingma last updated on 07/Feb/23
Answered by mr W last updated on 07/Feb/23
![=∫_0 ^(π/4) (dx/(2−(1−2 sin^2 x))) =∫_0 ^(π/4) (dx/(2−cos 2x)) =(1/2)∫_0 ^(π/2) (dx/(2−cos x)) =(1/( (√3)))[tan^(−1) ((√3) tan (x/2))]_0 ^(π/2) =(π/( 3(√3)))](Q186653.png)
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{2}−\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{2}−\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{2}−\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\:\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 08/Feb/23
![I=∫_0 ^(π/4) (dx/(1+2sin^2 x))=∫_0 ^(π/4) ((sec^2 x)/(sec^2 x+2tan^2 x))dx =∫_0 ^(π/4) ((sec^2 x)/(3tan^2 x+1))dx=∫_0 ^1 (dt/(3t^2 +1)) =(1/( (√3)))[arctan((√3)t)]_0 ^1 =(π/(3(√3)))](Q186656.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} {x}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{sec}^{\mathrm{2}} {x}+\mathrm{2tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{3tan}^{\mathrm{2}} {x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{arctan}\left(\sqrt{\mathrm{3}}{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$