What's the integration of 1/sinax?

Question

I tried it by using the substitution method (u = 2x) but the constant 'a' makes me reach a dead end when I divide the integrals to two parts. Now I've seen many videos on how to handle 1/sinx, but I can't reach a solution when the angle has another constant in it. What am I missing?

Answer 1

\begin{align*} \int \frac{1}{\sin(ax)} \,\mathrm{d}x &= \int \csc(ax) \,\mathrm{d}x \\ &= \frac{ -\ln | \csc(ax) + \cot(ax)|}{a} + C \text{.} \end{align*} If you are familiar with the antiderivative of cosecant, \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \cot x &= -\csc^2 x \\ \frac{\mathrm{d}}{\mathrm{d}x} \csc x &= -\csc x \cot x \end{align*} so $$ \frac{\mathrm{d}}{\mathrm{d}x} (\csc x + \cot x) = \csc x(\csc x + \cot x) $$ Then $$ - \frac{\frac{\mathrm{d}}{\mathrm{d}x} (\csc x + \cot x)}{(\csc x + \cot x)} = \csc x \text{.} $$ Antidifferentiating both sides and recognizing the left-hand side is a logarithm produces the result used above.