Trigonometric integrals $\int \sec^{m}(x)\tan^n(x)dx$
To integrate $\sec^{m}(x)\tan^{n}(x)$, we consider the following cases:
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If $m$ is even and nonzero, then we will use the substitution rule by letting $u=\tan(x)$ (so $du=\sec^2(x)dx$) and the trig identity $\sec^2(x) = \tan^2(x)+1$ to replace trig functions in terms of $u$. By doing so, integrals of trig functions becomes integrals of polynomials.
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If $n$ is odd, then we will use the substitution rule by letting $u=\sec(x)$ (so $du=\sec(x)\tan(x)dx$) and the trig identity $\tan^2(x) =\sec^2(x)-1$ to replace trig functions in terms of $u$. By doing so, integrals of trig functions becomes integrals of polynomials.
| m\n | $n$ is odd | $n$ is even |
|---|---|---|
| $m$ is odd | use 2. | |
| $m$ is even and $>0$ | use 1. or use 2. | use 1. |
The remanining cases:
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Case 1 ($m=0$) We will use the trig identity $\tan^2(x)=\sec^2(x)-1$, so $\int\tan^2k(x)dx$ becomses \(\int\tan^{2(k-1)}(x)(\sec^2(x)-1)dx = \int\tan^{2(k-1)}(x)\sec^2(x) - \int\tan^{2(k-1)}.\) The first part of the integral belongs the the case where $n$ is even and nonzero, and the second part has degree less than the previous integral. We will repeat this process untill we reduce the degree to $1$ or $0$.
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Case 2 ($m$ is odd and $n$ is even) We will use integration by part.
Trigonometric substitution
| pattern | trig identity | substitution |
|---|---|---|
| $a^2-x^2$ | $1-\sin^2(x) = \cos^2(x)$ ($1-\cos^2(x)=\sin^2(x)$) | $u=a\sin(x)$ ($u=a\cos(x)$) |
| $x^2 -a^2$ | $\tan^2(x) = \sec^2(x)-1$ | $u=a\sec(x)$ |
| $x^2+a^2$ | $\sec^2(x) = \tan^2(x)+1$ | $u=a\tan(x)$ |
Complete Square
Whenever we see somthing like $x^2+2ax$ in an integral, and substitution is not applicable. Try to complete the square.