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General antiderivative of a csc-cot and reciprocal power term

Find the general antiderivative of g(x)=2csc⁡(x)cot⁡(x)−6x.g(x) = 2\csc(x)\cot(x) - \dfrac{6}{x}.

Hint

Recall that ddx[csc⁡(x)]=−csc⁡(x)cot⁡(x)\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x) and ddx[ln⁡∣x∣]=1x\frac{d}{dx}[\ln|x|] = \frac{1}{x}. Reverse each derivative rule term by term, and don't forget the constant of integration.

Step-by-step solution

Split the antiderivative into two pieces using linearity: ∫2csc⁡(x)cot⁡(x) dx\int 2\csc(x)\cot(x)\,dx and ∫−6x dx\int -\dfrac{6}{x}\,dx. For the first piece, recall that ddx[−csc⁡(x)]=csc⁡(x)cot⁡(x)\frac{d}{dx}[-\csc(x)] = \csc(x)\cot(x), so ∫csc⁡(x)cot⁡(x) dx=−csc⁡(x)\int \csc(x)\cot(x)\,dx = -\csc(x), giving ∫2csc⁡(x)cot⁡(x) dx=−2csc⁡(x)\int 2\csc(x)\cot(x)\,dx = -2\csc(x).

For the second piece, use ∫1x dx=ln⁡∣x∣\int \frac{1}{x}\,dx = \ln|x|, so ∫−6x dx=−6ln⁡∣x∣\int -\dfrac{6}{x}\,dx = -6\ln|x|. Combining both results and adding the arbitrary constant gives the general antiderivative G(x)=−2csc⁡(x)−6ln⁡∣x∣+CG(x) = -2\csc(x) - 6\ln|x| + C.

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