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Question Number 213893 by issac last updated on 21/Nov/24

$${\int }_{-\pi }^{\pi }\frac{\mathrm{d}z}{1+{3\cos }^2\left(z\right)}=¿¿$$

Commented by BHOOPENDRA last updated on 21/Nov/24

$$\pi ???$$

Commented by issac last updated on 21/Nov/24

$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{why}\int \frac{\mathrm{d}r}{1+{3\cos }^2\left(r\right)}=\pi ??$$

Answered by mr W last updated on 21/Nov/24

$$=4{\int }_0^{\frac{\pi }{2}}\frac{dz}{1+3{\cos }^{2}z}$$$$=4{\int }_0^{\frac{\pi }{2}}\frac{dz}{4-3{\sin }^{2}z}$$$$=\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{dz}{{\left(\frac{2}{\sqrt{3}}\right)}^2-{\sin }^{2}z}$$$$=2{\left[{\tan }^{-1}\left(\frac{\tan z}{2}\right)\right]}_0^{\frac{\pi }{2}}$$$$=2\times \frac{\pi }{2}=\pi$$

Commented by issac last updated on 21/Nov/24

$$ohurgeniusthx:)$$

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