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

Answered by Frix last updated on 15/Dec/23

$$I=4\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{1+{3\sin }^{2}x}\stackrel{t=\tan x}{=}4\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{4t^2+1}=$$$$=2{\left[{\tan }^{-1}2t\right]}_0^{\mathrm{\infty }}=\pi$$

Answered by Mathspace last updated on 18/Dec/23

$$I={\int }_0^{2\pi }\frac{dx}{1+3\frac{1-cos\left(2x\right)}{2}}(2x=t)$$$$={\int }_0^{4\pi }\frac{dt}{2(1+\frac{3-3cost}{2})}$$$$={\int }_0^{4\pi }\frac{dt}{2+3-3cost}={\int }_0^{4\pi }\frac{dt}{5-3cost}$$$$={\int }_0^{2\pi }\frac{dt}{5-3cost}+{\int }_{2\pi }^{4\pi }\frac{dt}{5-3cost}(t=2\pi +\alpha )$$$$=2{\int }_0^{2\pi }\frac{dt}{5-3cost}(z=e^{it})$$$$=2{\int }_{\mid z\mid =1}\frac{dz}{iz(5-3\frac{z+z^{-1}}{2})}$$$$={\int }_{\mid z\mid =1}\frac{-4idz}{z(10-3z-3z^{-1})}$$$$={\int }_{\mid z\mid =1}\frac{-4i}{-3z^2+10z-3}dz$$$$={\int }_{\mid z\mid =1}\frac{4i}{3z^2-10z+3}dz$$$$let\mathrm{\Phi }\left(z\right)=\frac{4i}{3z^2-10z+3}?poles$$$${\mathrm{\Delta }}^{{}^{\prime }}={(-5)}^2-9=16\Rightarrow$$$$z_1=\frac{5+4}{3}=3and{z}_2=\frac{5-4}{3}=\frac{1}{3}$$$$\mid z_1\mid =3>1\left(totrow\right)$$$$\mid z_2\mid <1so$$$${\int }_{\mid z\mid =1}\mathrm{\Phi }\left(z\right)dz=2i\pi Res(\mathrm{\Phi },z_2)$$$$=2i\pi \times \frac{4i}{3(z_2-z_1)}=\frac{-8\pi }{3(\frac{1}{3}-3)}$$$$=\frac{-8\pi }{1-9}=\frac{-8\pi }{-8}=\pi \Rightarrow$$$$★{\int }_0^{2\pi }\frac{dx}{1+3{sin}^{2}x}=\pi ★$$

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