Mathematics

Solve:$\displaystyle \int_0^{2\pi}\dfrac{\cos x }{1+\sin^2 x} dx$

SOLUTION
$\int _{ 0 }^{ 2\pi }{ \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx } \\ Let\quad \sin x=t\\ \Rightarrow \cos xdx=dt\\ \because \int { \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx=\int { \dfrac { dt }{ 1+{ t }^{ 2 } } } } \\ ={ \tan }^{ -1 }t\\ ={ \tan }^{ -1 }(\sin x)\\ \therefore \int _{ 0 }^{ 2\pi }{ \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx } ={ \left[ { \tan }^{ -1 }(\sin x) \right] }_{ 0 }^{ 2\pi }\\ ={ \tan }^{ -1 }(\sin2\pi )-{ \tan }^{ -1 }(\sin0)\\ =0-0\\ =0$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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