Primitive of Secant Function/Secant plus Tangent Form/Proof 2
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Theorem
- $\ds \int \sec x \rd x = \ln \size {\sec x + \tan x} + C$
where $\sec x + \tan x \ne 0$.
Proof
| \(\ds \int \sec x \rd x\) | \(=\) | \(\ds \int \frac 1 {\cos x} \rd x\) | Secant is Reciprocal of Cosine |
We make the Weierstrass Substitution:
| \(\ds u\) | \(=\) | \(\ds \tan \frac x 2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \cos x\) | \(=\) | \(\ds \frac {1 - u^2} {1 + u^2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d u}\) | \(=\) | \(\ds \frac 2 {u^2 + 1}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \frac 1 {\cos x} \rd x\) | \(=\) | \(\ds \int \frac {1 + u^2} {1 - u^2} \frac 2 {u^2 + 1} \rd u\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \int \frac 1 {1 - u^2} \rd u\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \ln \size {\frac {1 + u} {1 - u} } + C\) | Primitive of $\dfrac 1 {a^2 - u^2}$: Logarithm Form | |||||||||||
| \(\ds \) | \(=\) | \(\ds \ln \size {\frac {1 + \tan \frac x 2} {1 - \tan \frac x 2} } + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \ln \size {\sec x + \tan x} + C\) | One Plus Tangent Half Angle over One Minus Tangent Half Angle |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration: Useful substitutions: Example $2$.