1 Use Integration by Parts on \int \sec^{3}x \, dx∫sec
3
xdx.
Let u=\sec{x}u=secx, dv=\sec^{2}xdv=sec
2
x, du=\sec{x}\tan{x} \, dxdu=secxtanxdx, v=\tan{x}v=tanx
2 Substitute the above into uv-\int v \, duuv−∫vdu.
\sec{x}\tan{x}-\int \tan^{2}x\sec{x} \, dxsecxtanx−∫tan
2
xsecxdx
3 Use Pythagorean Identities: \tan^{2}x=\sec^{2}x-1tan
2
x=sec
2
x−1.
\sec{x}\tan{x}-\int (\sec^{2}x-1)\sec{x} \, dxsecxtanx−∫(sec
2
x−1)secxdx
4 Expand (\sec^{2}x-1)\sec{x}(sec
2
x−1)secx.
\sec{x}\tan{x}-\int \sec^{3}x-\sec{x} \, dxsecxtanx−∫sec
3
x−secxdx
5 Use Sum Rule: \int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx.
\sec{x}\tan{x}-\int \sec^{3}x \, dx+\int \sec{x} \, dxsecxtanx−∫sec
3
xdx+∫secxdx
6 Set it as equal to the original integral \int \sec^{3}x \, dx∫sec
3
xdx.
\int \sec^{3}x \, dx=\sec{x}\tan{x}-\int \sec^{3}x \, dx+\int \sec{x} \, dx∫sec
3
xdx=secxtanx−∫sec
3
xdx+∫secxdx
7 Add \int \sec^{3}x \, dx∫sec
3
xdx to both sides.
\int \sec^{3}x \, dx+\int \sec^{3}x \, dx=\sec{x}\tan{x}+\int \sec{x} \, dx∫sec
3
xdx+∫sec
3
xdx=secxtanx+∫secxdx
8 Simplify \int \sec^{3}x \, dx+\int \sec^{3}x \, dx∫sec
3
xdx+∫sec
3
xdx to 2\int \sec^{3}x \, dx2∫sec
3
xdx.
2\int \sec^{3}x \, dx=\sec{x}\tan{x}+\int \sec{x} \, dx2∫sec
3
xdx=secxtanx+∫secxdx
9 Divide both sides by 22.
\int \sec^{3}x \, dx=\frac{\sec{x}\tan{x}+\int \sec{x} \, dx}{2}∫sec
3
xdx=
2
secxtanx+∫secxdx
10 Original integral solved.
\frac{\sec{x}\tan{x}+\int \sec{x} \, dx}{2}
2
secxtanx+∫secxdx
11 Use Trigonometric Integration: the integral of \sec{x}secx is \ln{(\sec{x}+\tan{x})}ln(secx+tanx).
\frac{\sec{x}\tan{x}+\ln{(\sec{x}+\tan{x})}}{2}
2
secxtanx+ln(secx+tanx)
12 Add constant.
\frac{\sec{x}\tan{x}+\ln{(\sec{x}+\tan{x})}}{2}+C
2
secxtanx+ln(secx+tanx)
+C