Question-201898

Question Number 201898 by sonukgindia last updated on 15/Dec/23

Answered by Frix last updated on 15/Dec/23

I=4∫_0 ^(π/2) (dx/(1+3sin^2 x)) =^(t=tan x) 4∫_0 ^∞ (dt/(4t^2 +1))= =2[tan^(−1) 2t]_0 ^∞ =π

$${I}=\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:{x}}\:\overset{{t}=\mathrm{tan}\:{x}} {=}\:\mathrm{4}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\mathrm{2}\left[\mathrm{tan}^{−\mathrm{1}} \:\mathrm{2}{t}\right]_{\mathrm{0}} ^{\infty} =\pi \\ $$

Answered by Mathspace last updated on 18/Dec/23

I=∫_0 ^(2π) (dx/(1+3((1−cos(2x))/2))) (2x=t) =∫_0 ^(4π) (dt/(2(1+((3−3cost)/2)))) =∫_0 ^(4π) (dt/(2+3−3cost))=∫_0 ^(4π) (dt/(5−3cost)) =∫_0 ^(2π) (dt/(5−3cost)) +∫_(2π) ^(4π) (dt/(5−3cost))(t=2π +α) =2∫_0 ^(2π) (dt/(5−3cost)) (z=e^(it) ) =2∫_(∣z∣=1) (dz/(iz(5−3((z+z^(−1) )/2)))) =∫_(∣z∣=1) ((−4i dz)/(z(10−3z−3z^(−1) ))) =∫_(∣z∣=1) ((−4i)/(−3z^2 +10z−3))dz =∫_(∣z∣=1) ((4i)/(3z^2 −10z+3))dz let Φ(z)=((4i)/(3z^2 −10z +3)) ?poles Δ^′ =(−5)^2 −9 =16 ⇒ z_1 =((5+4)/3)=3 and z_2 =((5−4)/3)=(1/3) ∣z_1 ∣=3 >1 (to trow) ∣z_2 ∣<1 so ∫_(∣z∣=1) Φ(z)dz=2iπ Res(Φ,z_2 ) =2iπ×((4i)/(3(z_2 −z_1 )))=((−8π)/(3((1/3)−3))) =((−8π)/(1−9))=((−8π)/(−8))=π ⇒ ★∫_0 ^(2π) (dx/(1+3sin^2 x))=π★

$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+\mathrm{3}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}\:\:\left(\mathrm{2}{x}={t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{4}\pi} \frac{{dt}}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{3}−\mathrm{3}{cost}}{\mathrm{2}}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{4}\pi} \frac{{dt}}{\mathrm{2}+\mathrm{3}−\mathrm{3}{cost}}=\int_{\mathrm{0}} ^{\mathrm{4}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{3}{cost}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{3}{cost}}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{3}{cost}}\left({t}=\mathrm{2}\pi\:+\alpha\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{3}{cost}}\:\:\left({z}={e}^{{it}} \right) \\ $$$$=\mathrm{2}\int_{\mid{z}\mid=\mathrm{1}} \frac{{dz}}{{iz}\left(\mathrm{5}−\mathrm{3}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{4}{i}\:{dz}}{{z}\left(\mathrm{10}−\mathrm{3}{z}−\mathrm{3}{z}^{−\mathrm{1}} \right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{4}{i}}{−\mathrm{3}{z}^{\mathrm{2}} +\mathrm{10}{z}−\mathrm{3}}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \frac{\mathrm{4}{i}}{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{10}{z}+\mathrm{3}}{dz} \\ $$$${let}\:\Phi\left({z}\right)=\frac{\mathrm{4}{i}}{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{10}{z}\:+\mathrm{3}}\:?{poles} \\ $$$$\Delta^{'} =\left(−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{9}\:=\mathrm{16}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{5}+\mathrm{4}}{\mathrm{3}}=\mathrm{3}\:\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{5}−\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\mathrm{3}\:>\mathrm{1}\:\left({to}\:{trow}\right) \\ $$$$\mid{z}_{\mathrm{2}} \mid<\mathrm{1}\:{so} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\Phi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\Phi,{z}_{\mathrm{2}} \right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mathrm{4}{i}}{\mathrm{3}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}=\frac{−\mathrm{8}\pi}{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{3}\right)} \\ $$$$=\frac{−\mathrm{8}\pi}{\mathrm{1}−\mathrm{9}}=\frac{−\mathrm{8}\pi}{−\mathrm{8}}=\pi\:\Rightarrow \\ $$$$\bigstar\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+\mathrm{3}{sin}^{\mathrm{2}} {x}}=\pi\bigstar \\ $$

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