## anonymous 5 years ago Integrate csc2xdx

1. anonymous

$\int\limits_{}^{}\csc(2x)dx$

2. myininaya

so 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 $\int\limits_{}^{}\frac{-du}{u}=-\ln|u|+C=-\ln|\csc(t)+\cot(t)|+C$ So the one here should be done similarly

3. myininaya

$\int\limits_{}^{}\csc(2x) dx=\int\limits_{}^{}\csc(2x) \cdot \frac{\csc(2x)+\cot(2x)}{\csc(2x)+\cot(2x)} dx$ $\text{ let } u=\csc(2x)+\cot(2x) => du=(-2 \csc(2x) \cot(2x)-2 \csc^2(2x)) dx$ $=>du=-2 \csc(2x)(\cot(2x)+\csc(2x))dx$ so we have $\int\limits_{}^{}-\frac{du}{2u}=\frac{-1}{2} \ln|u|+C=\frac{-1}{2}\ln|\csc(2x)+\cot(2x)|+C$

4. myininaya

so 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