Evaluate the indefinite integral integrate of arcsec(x)dx
- print Print
- list Cite
Expert Answers
mathsworkmusic
| Certified Educator
calendarEducator since 2012
write511 answers
starTop subjects are Math, Science, and Business
You need to integrate by parts
`int u (dv)/dx dx = uv - int v (du)/dx dx `
Let `v = x implies (dv)/dx = 1`
And `u = arcsecx` `implies (du)/dx = 1/(|x| sqrt(x^2 -1))` (see wikipedia page below)
Then `int arcsecx = xarcsecx - int x/(|x|sqrt(x^2-1)) dx`
Using integral of one over square root of a binomial and checking the cases `x < -1` and `x>1` (the domain of `x`)
`implies \int arcsecx = xarcsecx - ln|x + sqrt(x^2 -1)| + C` answer
Related Questions
- Evaluate the indefinite integral integrate of sin^4(x)cos^4(x)dx
- 1 Educator Answer
- Evaluate the indefinite integral integrate of (x^3(ln(x))dx)
- 2 Educator Answers
- Evaluate the indefinite integral integrate of (x)(arcsin(x))dx
- 1 Educator Answer
- Evaluate the integral [ln(x)/(x) dx]
- 1 Educator Answer
- Evaluate the indefinite integral `int x(8x+7)^8 dx`
- 1 Educator Answer