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Question Number 200318 by sonukgindia last updated on 17/Nov/23

Answered by Sutrisno last updated on 17/Nov/23

$$={\int }_0^{\pi }\frac{\frac{1}{{cos}^{2}x}}{\frac{1}{{cos}^{2}x}+\frac{{cos}^{2}x}{{cos}^{2}x}}dx$$$$={\int }_0^{\pi }\frac{{sec}^{2}x}{{tan}^{2}x+2}dx$$$$misal=:tanx=\sqrt{2}tan\theta \to dx=\frac{\sqrt{2}{sec}^2\theta }{{sec}^{2}x}d\theta$$$$=\int \frac{{sec}^{2}x}{{\left(\sqrt{2}tan\theta \right)}^2+2}.\frac{\sqrt{2}{sec}^2\theta }{{sec}^{2}x}d\theta$$$$=\int \frac{\sqrt{2}{sec}^2\theta }{2{sec}^2\theta }d\theta$$$$=\frac{\sqrt{2}}{2}\int d\theta$$$$=\frac{\sqrt{2}}{2}\theta$$$$=\frac{\sqrt{2}}{2}arctan\left(\frac{tanx}{\sqrt{2}}\right){\mid }_0^{\pi }$$$$=0$$

Commented by mr W last updated on 17/Nov/23

$$\frac{1}{2}<\frac{1}{1+{\cos }^{2}x}<1$$$$\Rightarrow \frac{\pi }{2}<{\int }_0^{\pi }\frac{dx}{1+{\cos }^{2}x}<\pi$$$$thatmeans{\int }_0^{\pi }\frac{dx}{1+{\cos }^{2}x}canneverbe0.$$

Answered by MM42 last updated on 17/Nov/23

$$I=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{cos}^{2}x}=2{\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^{2}x}{2+{tan}^{2}x}dx$$$${tanx=u\Rightarrow I=2{\int }_0^{\mathrm{\infty }}\frac{du}{2+u^2}=\sqrt{2}{tan}^{-1}\left(\frac{u}{\sqrt{2}}\right)]}_0^{\mathrm{\infty }}$$$$=\frac{\sqrt{2}}{2}\pi ✓$$

Commented by MM42 last updated on 17/Nov/23

$$★★★Attention★★★$$$$I_7={\int }_0^{\pi }\frac{dx}{1+{cos}^{2}x}=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{cos}^{2}x}=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{sin}^2(\frac{\pi }{2}-x)}$$$$\frac{\pi }{2}-x=u\Rightarrow I_7=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{sin}^{2}x}=I_8$$$$$$

Commented by MM42 last updated on 17/Nov/23

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