Evaluate the indefinite trigonometric double angle integral

Question

The integral is $$\int \csc(2x)dx$$ I have tried transforming it into $$\int \frac{1}{\sin(2x)}dx $$ then using u-sub. I get $$u=\sin(2x)$$ $$du=2\cos(2x)$$ but I have trouble getting $du=1$, because then I can take the $\ln(u)$

I then thought of using one of $\cos(2x)$'s double angle formulas but still nothing.

To help get the proper solution, I looked up the answer in the back of the book and its $$-\frac{1}{2} \ln\mid\csc(2x)+\cot(2x)\mid+C$$

Answer 1

$$\int \csc(2x)dx=\int \sec\left(\frac{\pi}{2}-2x\right)dx$$ Now, taking $u=\frac{\pi}{2}-2x$ we get $$-\frac{1}{2} \int \sec(u)du$$ and recall that, this is equal to $$-\frac{1}{2} \log \big| \sec(u)+\tan(u)\big|+c$$ Returning the variables $$-\frac{1}{2} \log \left| \sec\left(\frac{\pi}{2}-2x\right)+\tan\left(\frac{\pi}{2}-2x\right)\right|+c=-\frac{1}{2}\log \big|\csc(2x)+\cot(2x)\big|+c$$

Answer 2

$$\int \csc(2x)dx=\int \frac{1}{\sin(2x)}dx=\int \frac{1}{2\sin(x) \cos(x)}dx=$$

$$\int \frac{\sin(x)}{2\sin^2(x) \cos(x)}dx=$$

$$\int \frac{\sin(x)}{2(1-\cos^2(x)) \cos(x)}dx$$

The substitution $u=\cos(x)$, $du=-\sin(x)dx$ changes the integral into

$$\int \frac{du}{2u(u^2-1)}$$

which is solved by partial fractions.

Answer 3

As,

$$\int\csc(ax)dx = - \frac{1}{a}\ln|\csc(ax)+\cot(ax)|+c$$

$$\color{green}{\int\csc(2x)dx = - \frac{1}{2}\ln|\csc(2x)+\cot(2x)|+c}$$

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If you need a proof for the above formula,

$$\int\csc(ax)dx = \int\frac{\csc(ax)\big[\csc(ax)+\cot(ax)\big]dx}{\csc(ax)+\cot(ax)}$$

Let $$t= \csc(ax)+\cot(ax)\implies dt = -\bigg[a\csc(ax)\cot(ax)+a\csc^2(ax)\bigg]dx $$ $$\csc(ax)[\csc(ax)+\cot(ax)\big]dx = \frac{-dt}{a}$$

So, $$\color{blue}{\int\csc(ax)dx = -\frac{1}{a}\int\frac{dt}{t} = -\frac{1}{a}\ln|t| +c= -\frac{1}{a}\ln|\csc(ax)+\cot(ax)|+c}$$

Answer 4

Here we will address the integral: \begin{equation} I = \int \csc(2x)\:dx \end{equation} First let $u = 2x$: \begin{equation} I = \int \csc(u)\cdot \frac{1}{2}\:du = \frac{1}{2} \int \frac{1}{\sin(u)}\:du \end{equation} We now employ the Weirerstrauss Substitution $t = \tan\left(\frac{u}{2}\right)$: \begin{equation} I =\frac{1}{2} \int \frac{1}{\frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt = \frac{1}{2} \int \frac{1}{t}\:dt = \frac{1}{2}\ln\left| t\right| + C = \frac{1}{2}\ln\left|\tan\left(\frac{u}{2}\right) \right| + C = \frac{1}{2}\ln\left|\tan\left(x \right) \right| + C \end{equation} Where $C$ is the constant of integration.