求证积分$I_n = \int \tanh^{n}x$的递推公式$I_n = I_{n-2} - \frac{1}{n-1}\tanh^{n-1}x$
No worries, let's walk through this step by step—it's all about cleverly rewriting the integrand and using basic integration techniques.
Rewrite the integrand to link to lower powers
Start by breaking down $\tanh^n x$ into a product of a lower power of $\tanh x$ and $\tanh^2 x$:
$$\tanh^n x = \tanh^{n-2} x \cdot \tanh^2 x$$
We know from hyperbolic identities that $\tanh^2 x = 1 - \text{sech}^2 x$, so substitute that in:
$$\tanh^n x = \tanh^{n-2} x (1 - \text{sech}^2 x)$$Split the integral into two parts
Now expand the integral using this rewritten form:
$$I_n = \int \tanh^n x dx = \int \tanh^{n-2} x (1 - \text{sech}^2 x) dx = \int \tanh^{n-2} x dx - \int \tanh^{n-2} x \cdot \text{sech}^2 x dx$$
The first integral here is exactly $I_{n-2}$, so we can rewrite this as:
$$I_n = I_{n-2} - \int \tanh^{n-2} x \cdot \text{sech}^2 x dx$$Evaluate the remaining integral with substitution
Let's focus on the integral $\int \tanh^{n-2} x \cdot \text{sech}^2 x dx$. Use the substitution $u = \tanh x$—notice that the derivative of $\tanh x$ is $\text{sech}^2 x$, so:
$$du = \text{sech}^2 x dx$$
Substitute into the integral:
$$\int u^{n-2} du = \frac{u^{n-1}}{n-1} + C = \frac{\tanh^{n-1} x}{n-1} + C$$Combine everything to get the recurrence relation
Plug this result back into our earlier expression for $I_n$:
$$I_n = I_{n-2} - \frac{1}{n-1}\tanh^{n-1} x + C$$
Since $I_{n-2}$ already includes the constant of integration (as it's an indefinite integral), we can omit the separate constant term, leaving the final recurrence:
$$\boxed{I_n = I_{n-2} - \frac{1}{n-1}\tanh^{n-1} x}$$
内容的提问来源于stack exchange，提问作者m.roussev
