Question

Indefinite integral of sec 2x + tan2x dx

Answer 1

Assuming you mean
`\sec2x+\tan2x`, you have


`\displaystyle\int(\sec2x+\tan2x)\,\mathrm dx=\int\sec2x(\sec2x+\tan2x)/(\sec2x+\tan2x)\,\mathrm dx+\int(\sin2x)/(\cos2x)\,\mathrm dx`

`\displaystyle=\int(\sec^22x+\sec2x\tan2x)/(\sec2x+\tan2x)\,\mathrm dx+\int(\sin2x)/(\cos2x)\,\mathrm dx`

For the first integral, set
`u=\sec2x+\tan2x`, so that
`\mathrm du=2(\sec2x\tan2x+\sec^22x)\,\mathrm dx`; for the second,
`v=\cos2x`, so that
`\mathrm dv=-2\sin2x\,\mathrm dx`. Then you have


`\displaystyle\frac12\int\frac{\mathrm du}u-\frac12\int\frac{\mathrm dv}v`

`\displaystyle=\frac12(\ln|u|-\ln|v|)+C`

`\displaystyle=\ln\left|\frac uv\right|^(1/2)+C`

`\displaystyle=\ln\left|(\sec2x+\tan2x)/(\cos2x)\right|^(1/2)+C`

`\displaystyle=\ln\left|\sec^22x(1+\sin2x)\right|^(1/2)+C`