Evaluate; integral of sec³x dx

Mathematics

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Answer

12(sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣)+C\frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C

Step-by-step solution

9 steps

  1. Use integration by parts.

    Let u=sec⁡xu = \sec x and dv=sec⁡2x dxdv = \sec^2 x \, dx.
    Then du=sec⁡xtan⁡x dxdu = \sec x \tan x \, dx and v=tan⁡xv = \tan x.

  2. Apply the integration by parts formula

    ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

    ∫sec⁡3x dx=sec⁡xtan⁡x−∫tan⁡x(sec⁡xtan⁡x) dx\int \sec^3 x \, dx = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx ∫sec⁡3x dx=sec⁡xtan⁡x−∫sec⁡xtan⁡2x dx\int \sec^3 x \, dx = \sec x \tan x - \int \sec x \tan^2 x \, dx

  3. Use the identity tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1.

    ∫sec⁡3x dx=sec⁡xtan⁡x−∫sec⁡x(sec⁡2x−1) dx\int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx ∫sec⁡3x dx=sec⁡xtan⁡x−∫(sec⁡3x−sec⁡x) dx\int \sec^3 x \, dx = \sec x \tan x - \int (\sec^3 x - \sec x) \, dx

  4. Distribute the integral.

    ∫sec⁡3x dx=sec⁡xtan⁡x−∫sec⁡3x dx+∫sec⁡x dx\int \sec^3 x \, dx = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx

  5. Let I=∫sec⁡3x dxI = \int \sec^3 x \, dx.

    The equation becomes:

    I=sec⁡xtan⁡x−I+∫sec⁡x dxI = \sec x \tan x - I + \int \sec x \, dx

  6. Add II to both sides.

    2I=sec⁡xtan⁡x+∫sec⁡x dx2I = \sec x \tan x + \int \sec x \, dx

  7. Evaluate the integral of sec⁡x\sec x.

    ∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C\int \sec x \, dx = \ln|\sec x + \tan x| + C

  8. Substitute this back into the equation for $2I$.

    2I=sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣+C2I = \sec x \tan x + \ln|\sec x + \tan x| + C

  9. Solve for II.

    I=12(sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣)+CI = \frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C

    The final answer is 12(sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣)+C\frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C.

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