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How do you integrate $ \int \csc^{3}x\,dx $ ?

I know that you have to change the integral into $\int \csc x (1 + \cot^2 x)\,dx$

But what do you do next?

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how about $ \int \csc ^3 x \, dx = \int \frac 1{\sin ^3 x} \, dx = \int \frac{\sin x \, dx}{\sin ^4 x} = - \int \frac{du}{(1-u^2)^2}$ where $u = \cos x.$

we can now use partial fraction to decompose $$ \frac 1{(u^2 -1)^2} = \frac{1}{(u-1)^2(u+1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{u+1}+\frac{D}{(u+1)^2}\tag 1$$ you find that $B = D = \frac14.$ find the constants $A, C.$ now you can integrate every term in $(1).$

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$$\int\csc^3(x)\space\text{d}x=$$


Use the reduction formula, where $m=3$:

$$\int \csc^m(x)\space\text{d}x=-\frac{\cos(x)\csc^{m-1}(x)}{m-1}+\frac{m-2}{m-1}\int\csc^{m-2}(x)\space\text{d}x$$


$$-\frac{\cot(x)\csc(x)}{2}+\frac{1}{2}\int\csc(x)\space\text{d}x=$$


Multiply numerator and denominator of $\csc(x)$ by $\cot(x)+\csc(x)$:


$$-\frac{\cot(x)\csc(x)}{2}+\frac{1}{2}\int-\frac{-\cot(x)\csc(x)-\csc^2(x)}{\cot(x)+\csc(x)}\space\text{d}x=$$


Substitute $u=\cot(x)+\csc(x)$ and $\text{d}u=(-\csc^2(x)-\cot(x)\csc(x))\space\text{d}x$:


$$-\frac{\cot(x)\csc(x)}{2}-\frac{1}{2}\int\frac{1}{u}\space\text{d}u=$$ $$-\frac{\cot(x)\csc(x)}{2}-\frac{\ln\left|u\right|}{2}+\text{C}=$$ $$-\frac{\cot(x)\csc(x)}{2}-\frac{\ln\left|\cot(x)+\csc(x)\right|}{2}+\text{C}=$$ $$-\frac{\cot(x)\csc(x)-\ln\left|\cot(x)+\csc(x)\right|}{2}+\text{C}$$

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HINT: $$\sin(x)^3=\sin(x)^2\sin(x)=(1-\cos(x)^2)\sin(x)$$ setting $t=\sin(x)$ we get $$dx=-\frac{dt}{\sin(x)}$$ and our integral will be $$-\int \frac{dt}{(1-t^2)\sin(x)^2}=-\int \frac{dt}{(1-t^2)^2}$$

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