Question Number 200894 by sonukgindia last updated on 26/Nov/23
Answered by Frix last updated on 26/Nov/23
![=4∫_0 ^(π/2) (dx/(1+4sin^2 x)) =^(t=tan x) =4∫_0 ^∞ (dt/(5t^2 +1))=(4/( (√5)))[tan^(−1) (√5)t]_0 ^∞ =((2π)/( (√5)))](https://www.tinkutara.com/question/Q200907.png)
$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{4sin}^{\mathrm{2}} \:{x}}\:\overset{{t}=\mathrm{tan}\:{x}} {=} \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}\left[\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{5}}{t}\right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{5}}} \\ $$
Answered by Calculusboy last updated on 28/Nov/23
![Solution: multiply numerator and denominator by ((1/(cos^2 x))) I=∫_0 ^(𝛑/2) ((1/(cos^2 x))/(1/(cos^2 x+((4sin^2 x)/(cos^2 x)))))dx Nb: tanx=((sinx)/(cosx)) and secx=(1/(cosx)) I=∫_0 ^(π/2) ((sec^2 x)/(sec^2 x+4tan^2 x))dx recall that sec^2 x=1+tan^2 x I=∫_0 ^(𝛑/2) ((sec^2 x)/(1+tan^2 x+4tan^2 x))dx let u=tanx dx=(du/(sec^2 x)) when x=(𝛑/2) u=∞ and when x=0 u=0 I=∫_0 ^∞ ((sec^2 x)/(1+5u^2 ))∙(du/(sec^2 x)) ⇔ I=(1/5)∫_0 ^∞ (du/((1/5)+u^2 )) I=(1/5)∫_0 ^∞ (du/(u^2 +((1/( (√5))))^2 )) ⇔ I=(1/5)×(1/(((1/( (√5))))))[tan^(−1) ((x/(((1/( (√5)))))))]_0 ^∞ +C I=((√5)/5)[tan^(−1) (∞)−tan^(−1) (0)] I=((√5)/5)[(𝛑/2)−0] I=((𝛑(√5))/(10))](https://www.tinkutara.com/question/Q201019.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{multiply}}\:\boldsymbol{{numerator}}\:\boldsymbol{{and}}\:\boldsymbol{{denominator}} \\ $$$$\boldsymbol{{by}}\:\left(\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}\right) \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}}{\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}+\frac{\mathrm{4}\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}}}\boldsymbol{{dx}}\:\boldsymbol{{Nb}}:\:\boldsymbol{{tanx}}=\frac{\boldsymbol{{sinx}}}{\boldsymbol{{cosx}}}\:\boldsymbol{{and}}\:\boldsymbol{{secx}}=\frac{\mathrm{1}}{\boldsymbol{{cosx}}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}+\mathrm{4}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\:\boldsymbol{{recall}}\:\boldsymbol{{that}}\:\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}=\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}+\mathrm{4}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\:\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{tanx}}\:\:\:\boldsymbol{{dx}}=\frac{\boldsymbol{{du}}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{{u}}=\infty\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{u}}=\mathrm{0} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\mathrm{1}+\mathrm{5}\boldsymbol{{u}}^{\mathrm{2}} }\centerdot\frac{\boldsymbol{{du}}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{du}}}{\frac{\mathrm{1}}{\mathrm{5}}+\boldsymbol{{u}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{du}}}{\boldsymbol{{u}}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} }\:\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)}\left[\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{x}}}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)}\right)\right]_{\mathrm{0}} ^{\infty} +\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\left[\boldsymbol{{tan}}^{−\mathrm{1}} \left(\infty\right)−\boldsymbol{{tan}}^{−\mathrm{1}} \left(\mathrm{0}\right)\right] \\ $$$$\boldsymbol{{I}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\left[\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right] \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}\sqrt{\mathrm{5}}}{\mathrm{10}} \\ $$$$ \\ $$