Cao Yi

Indefinite Integrals of $\tan^n(x)$ and $\cot^n(x)$

Index

In this article, we derive the reduction formulas for $\int \tan^n(x) \,\mathrm{d}x$ and $\int \cot^n(x) \,\mathrm{d}x$.

We will use the following identities and basic formulas.

\[\begin{align} &\tan^2(x) + 1 = \sec^2(x), \quad \cot^2(x) + 1 = \csc^2(x)\\ &\dfrac{\mathrm{d}}{\mathrm{d}x}\tan(x) = \sec^2(x), \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x) = -\csc^2(x)\\ &\int \sec^2(x) \,\mathrm{d}x = \tan(x) + C, \quad \int \csc^2(x) \,\mathrm{d}x = -\cot(x) + C \end{align}\]

tan

Let $I_n = \int \tan^n(x) \,\mathrm{d}x$. Then

\[\begin{aligned} I_0 &= \int \,\mathrm{d}x = x + C \\ I_1 &= \int \tan(x) \,\mathrm{d}x = -\ln(\vert \cos(x) \vert) + C = \ln(\vert \sec(x) \vert) + C \end{aligned}\]

Let $n$ be an integer with $n>1$. Then

\[\begin{aligned} \int \tan^n(x) \mathrm{d}x &= \int \tan^{n-2}(x)\cdot\tan^2(x) \,\mathrm{d}x \\ &= \int \tan^{n-2}(x)\cdot(\sec^2(x)-1) \,\mathrm{d}x \\ &= \int \tan^{n-2}(x)\cdot\sec^2(x) \,\mathrm{d}x - \int \tan^{n-2}(x) \,\mathrm{d}x \\ &= \int \tan^{n-2}(x) \,\mathrm{d}\tan(x) - \int \tan^{n-2}(x) \,\mathrm{d}x \\ &= \dfrac{\tan^{n-1}(x)}{n-1} - \int \tan^{n-2}(x) \,\mathrm{d}x \end{aligned}\]

Thus, we obtain the reduction formula

\[I_n = \dfrac{\tan^{n-1}(x)}{n-1} - I_{n-2}, \qquad n > 1\]

cot

Let $I_n = \int \cot^n(x) \,\mathrm{d}x$. Then

\[\begin{aligned} I_0 &= \int \,\mathrm{d}x = x + C \\ I_1 &= \int \cot(x) \,\mathrm{d}x = \ln(\vert \sin(x) \vert) + C = -\ln(\vert \csc(x) \vert) + C \end{aligned}\]

Let $n$ be an integer with $n>1$. Then

\[\begin{aligned} \int \cot^n(x) \mathrm{d}x &= \int \cot^{n-2}(x)\cdot\cot^2(x) \,\mathrm{d}x \\ &= \int \cot^{n-2}(x)\cdot(\csc^2(x)-1) \,\mathrm{d}x \\ &= \int \cot^{n-2}(x)\cdot\csc^2(x) \,\mathrm{d}x - \int \cot^{n-2}(x) \,\mathrm{d}x \\ &= -\int \cot^{n-2}(x) \,\mathrm{d}\cot(x) - \int \cot^{n-2}(x) \,\mathrm{d}x \\ &= -\dfrac{\cot^{n-1}(x)}{n-1} - \int \cot^{n-2}(x) \,\mathrm{d}x \end{aligned}\]

Thus, we obtain the reduction formula

\[I_n = -\dfrac{\cot^{n-1}(x)}{n-1} - I_{n-2}, \qquad n > 1\]