Question Number 213893 by issac last updated on 21/Nov/24 | ||
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$$\int_{\:−\pi} ^{\:\:\pi} \:\:\frac{\mathrm{d}{z}}{\mathrm{1}+\mathrm{3cos}^{\mathrm{2}} \left({z}\right)}=¿¿\:\:\: \\ $$ | ||
Commented by BHOOPENDRA last updated on 21/Nov/24 | ||
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$$\pi??? \\ $$ | ||
Commented by issac last updated on 21/Nov/24 | ||
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$$\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{why}\:\int\:\frac{\mathrm{d}{r}}{\mathrm{1}+\mathrm{3cos}^{\mathrm{2}} \left({r}\right)}=\pi\:?? \\ $$ | ||
Answered by mr W last updated on 21/Nov/24 | ||
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$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dz}}{\mathrm{1}+\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:{z}} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dz}}{\mathrm{4}−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{z}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dz}}{\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:{z}} \\ $$$$=\mathrm{2}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{z}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{2}×\frac{\pi}{\mathrm{2}}=\pi \\ $$ | ||
Commented by issac last updated on 21/Nov/24 | ||
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$$\left.{oh}\:{u}\:{r}\:{genius}\:{thx}\::\right) \\ $$ | ||