Yesterday, while studying an optimization problem, I encountered an integral like this: \int \frac{1}{\cos^3 \theta} d\theta
Then I studied the integration of this type of function. Generally, these integrals can be written as \int \cos^n \theta d\theta or \int \sin^n \theta d\theta, where n is an integer.
First, let’s solve the case n=1. We easily have \int \cos \theta d\theta = \sin \theta + C or \int \sin \theta d\theta = -\cos \theta + C, which are basic results.
If n is a positive integer greater than 1, we can use a recurrence method: \begin{aligned} \int \cos^n \theta d\theta &= \int \cos^{n-1} \theta d(\sin\theta) \\ &= \cos^{n-1} \theta \sin\theta - \int \sin\theta d(\cos^{n-1}\theta) \\ &= \cos^{n-1} \theta \sin\theta + (n-1)\int \sin^2\theta \cos^{n-2}\theta d\theta \\ &= \cos^{n-1} \theta \sin\theta + (n-1)\int (1-\cos^2\theta) \cos^{n-2}\theta d\theta \\ &= \cos^{n-1} \theta \sin\theta + (n-1)\int \cos^{n-2}\theta d\theta - (n-1)\int \cos^n\theta d\theta \end{aligned} (Integration by parts was used here.)
That is, n\int \cos^n \theta d\theta = \cos^{n-1} \theta \sin\theta + (n-1)\int \cos^{n-2}\theta d\theta, which gives: \int \cos^n \theta d\theta = \frac{1}{n} \left[ \cos^{n-1} \theta \sin\theta + (n-1)\int \cos^{n-2}\theta d\theta \right] This completes the recurrence. There is a similar formula for sine, which I will write directly: \int \sin^n \theta d\theta = \frac{1}{n} \left[ -\sin^{n-1} \theta \cos\theta + (n-1)\int \sin^{n-2}\theta d\theta \right]
However, last night I thought about transforming it into a polynomial integral: \int \cos^n \theta d\theta = \int \cos^{n-1} \theta d(\sin\theta) If we let \sin\theta = x, the integral becomes: \int (1-x^2)^{\frac{n-1}{2}} dx
If n is odd (positive or negative), this integral is quite easy to handle. If it is even or another rational number, it requires the use of Hypergeometric Functions to express. Here, we only discuss the case where n is odd.
When n is positive, it can be integrated directly by polynomial expansion, so we will not discuss it further. When n is negative, it is equivalent to discussing the integral: \int \frac{1}{(1-x^2)^k} dx where k is a positive integer. When k=1, we have: \int \frac{1}{1-x^2} dx = \frac{1}{2} \ln \left| \frac{1+x}{1-x} \right|
When k > 1, let I_k = \int \frac{1}{(1-x^2)^k} dx. We have: \begin{aligned} I_k &= \int \frac{1-x^2+x^2}{(1-x^2)^k} dx = I_{k-1} + \int \frac{x^2}{(1-x^2)^k} dx \\ &= I_{k-1} - \frac{1}{2} \int \frac{x}{(1-x^2)^k} d(1-x^2) \\ &= I_{k-1} + \frac{1}{2(k-1)} \int x d[(1-x^2)^{1-k}] \\ &= I_{k-1} + \frac{1}{2(k-1)} \left[ x (1-x^2)^{1-k} - \int (1-x^2)^{1-k} dx \right] \\ &= \left( \frac{2k-3}{2k-2} \right) I_{k-1} + \frac{x}{2(k-1)(1-x^2)^{k-1}} \end{aligned} The recurrence is complete.
Here I directly write down the results I might use: \int \frac{1}{\cos^3 \theta} d\theta = \frac{1}{4} \left( \frac{2\sin \theta}{\cos^2 \theta} + \ln \left| \frac{1+\sin \theta}{1-\sin \theta} \right| \right) \int \frac{1}{\sin^3 \theta} d\theta = -\frac{1}{4} \left( \frac{2\cos \theta}{\sin^2 \theta} + \ln \left| \frac{1+\cos \theta}{1-\cos \theta} \right| \right)
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