Integral of \( \sec^4(x) \)

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Evaluate the integral \[ \int \sec^4 x \; dx \]
Write the integrand \( \sec^4 x \) as the product \( \sec^2 x \sec^2 x \)
\[ \int \sec^4 x \; dx = \int \sec^2 x \; \sec^2 x \; dx \]
Use the
trigonometric identity \( \sec^2 x = \tan^2 x + 1 \) to write the integral as follows \[ \int \sec^4 x \; dx = \int (\tan^2 x + 1) \sec^2 x \; dx \] Expand the integrand and rewrite the integral as a sum of integrals \[ \int \sec^4 x \; dx = \int \tan^2 x \sec^2 x \; dx + \int \sec^2 x \; dx \] Use Integration by Substitution : Let \( u = \tan x \) and hence \( \dfrac{du}{dx} = \sec^2 x \) or \( dx = \dfrac{1}{\sec^2 x} du \) to write \[ \int \sec^4 x \; dx = \int u^2 \sec^2 x \dfrac{1}{\sec^2 x} du + \int \sec^2 x \; dx \] Simplify \[ \int \sec^4 x \; dx = \int u^2 du + \int \sec^2 x \; dx \] Evaluate using the integral formulas \( \displaystyle \int u^2 du = (1/3) u^3 \) and the common integral \( \displaystyle \int \sec^2 x \; dx = \tan x\) to write \[ \int \sec^4 x \; dx = \dfrac{1}{3} u^3 + \tan x + c \] Substitute back \( \displaystyle u = \tan x \) to obtain the final answer \[ \boxed { \int \sec^4 x \; dx = \dfrac{1}{3} \tan^3 x + \tan x + c } \]



More References and Links

  1. Table of Integral Formulas
  2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
  3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
  4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8

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