∫sec⁡3x dx\int \sec^{3}x \, dx
1
Use Integration by Parts on ∫sec⁡3x dx\int \sec^{3}x \, dx.
Let u=sec⁡xu=\sec{x}, dv=sec⁡2xdv=\sec^{2}x, du=sec⁡xtan⁡x dxdu=\sec{x}\tan{x} \, dx, v=tan⁡xv=\tan{x}

2
Substitute the above into uv−∫v duuv-\int v \, du.
sec⁡xtan⁡x−∫tan⁡2xsec⁡x dx\sec{x}\tan{x}-\int \tan^{2}x\sec{x} \, dx

3
Use Pythagorean Identities: tan⁡2x=sec⁡2x−1\tan^{2}x=\sec^{2}x-1.
sec⁡xtan⁡x−∫(sec⁡2x−1)sec⁡x dx\sec{x}\tan{x}-\int (\sec^{2}x-1)\sec{x} \, dx

4
Expand.
sec⁡xtan⁡x−∫sec⁡3x−sec⁡x dx\sec{x}\tan{x}-\int \sec^{3}x-\sec{x} \, dx

5
Use Sum Rule: ∫f(x)+g(x) dx=∫f(x) dx+∫g(x) dx\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx.
sec⁡xtan⁡x−∫sec⁡3x dx+∫sec⁡x dx\sec{x}\tan{x}-\int \sec^{3}x \, dx+\int \sec{x} \, dx

6
Set it as equal to the original integral ∫sec⁡3x dx\int \sec^{3}x \, dx.
∫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

7
Add ∫sec⁡3x dx\int \sec^{3}x \, dx to both sides.
∫sec⁡3x dx+∫sec⁡3x dx=sec⁡xtan⁡x+∫sec⁡x dx\int \sec^{3}x \, dx+\int \sec^{3}x \, dx=\sec{x}\tan{x}+\int \sec{x} \, dx

8
Simplify  ∫sec⁡3x dx+∫sec⁡3x dx\int \sec^{3}x \, dx+\int \sec^{3}x \, dx  to  2∫sec⁡3x dx2\int \sec^{3}x \, dx.
2∫sec⁡3x dx=sec⁡xtan⁡x+∫sec⁡x dx2\int \sec^{3}x \, dx=\sec{x}\tan{x}+\int \sec{x} \, dx

9
Divide both sides by 22.
∫sec⁡3x dx=sec⁡xtan⁡x+∫sec⁡x dx2\int \sec^{3}x \, dx=\frac{\sec{x}\tan{x}+\int \sec{x} \, dx}{2}

10
Original integral solved.
sec⁡xtan⁡x+∫sec⁡x dx2\frac{\sec{x}\tan{x}+\int \sec{x} \, dx}{2}

11
Use Trigonometric Integration: the integral of sec⁡x\sec{x} is ln⁡(sec⁡x+tan⁡x)\ln{(\sec{x}+\tan{x})}.
sec⁡xtan⁡x+ln⁡(sec⁡x+tan⁡x)2\frac{\sec{x}\tan{x}+\ln{(\sec{x}+\tan{x})}}{2}

12
Add constant.
sec⁡xtan⁡x+ln⁡(sec⁡x+tan⁡x)2+C\frac{\sec{x}\tan{x}+\ln{(\sec{x}+\tan{x})}}{2}+C

Done
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