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anonymous
 4 years ago
Integrate csc2xdx
anonymous
 4 years ago
Integrate csc2xdx

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\csc(2x)dx\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3so recall \[\int\limits_{}^{}\csc(t) dt=\int\limits_{}^{}\csc(t) \cdot \frac{\csc(t)+\cot(t)}{\csc(t)+\cot(t)} dt\] \[\text{ let } u=\csc(t)+\cot(t) => du=(\csc(t) \cot(t)\csc^2(t) )dt=\csc(t)(\cot(t)+\csc(t)) dt\] so we have \[\int\limits_{}^{}\frac{du}{u}=\lnu+C=\ln\csc(t)+\cot(t)+C\] So the one here should be done similarly

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\int\limits_{}^{}\csc(2x) dx=\int\limits_{}^{}\csc(2x) \cdot \frac{\csc(2x)+\cot(2x)}{\csc(2x)+\cot(2x)} dx\] \[\text{ let } u=\csc(2x)+\cot(2x) => du=(2 \csc(2x) \cot(2x)2 \csc^2(2x)) dx\] \[=>du=2 \csc(2x)(\cot(2x)+\csc(2x))dx\] so we have \[\int\limits_{}^{}\frac{du}{2u}=\frac{1}{2} \lnu+C=\frac{1}{2}\ln\csc(2x)+\cot(2x)+C\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3so we can conclude if we have \[\int\limits_{}^{}\csc(ax) dx=\frac{1}{a}\ln\csc(ax)+\cot(ax)+C\] where a is a constant not equal to 0
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