Evaluate the indefinite integral integrate of (sec^2(t))/(sqrt(1-9tan^2(t)))dt
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
You should come up with the following substitution such that:
`tan t = u => 1/(cos^2t) dt = du => sec^2 t dt = du`
Changing the variable yields:
`int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = int (du)/(sqrt(1-9u^2))`
You should come up with the following substitution such that:
`u = (sin v)/3 => u^2 = (sin^2 v)/9`
`du = (cos v)/3 dv`
Changing the variable yields:
`int (du)/(sqrt(1-9u^2)) = int (cos v)/3/(sqrt(1-9((sin^2 v)/9))) dv`
`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv`
You need to use the fundamental formula of trigonometry such that:
`1 - sin^2 v = cos^2 v => sqrt(1 - sin^2 v) = sqrt(cos^2 v) = cos v`
`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) int cos v/cos v dv`
`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) int dv`
`(1/3) int cos v/(sqrt(1 - sin^2 v)) dv = (1/3) v + c`
Substituting back `arcsin (3u)` fot v yields:
`int (du)/(sqrt(1-9u^2)) = (1/3)arcsin (3u) + c`
Substituting back `tan t` for u yields:
`int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = (1/3) arcsin (3tan t) + c`
Hence, evaluating the given integral yields `int (sec^2 t)/(sqrt(1 - 9tan^2 t))dt = (1/3) arcsin (3tan t) + c.`
Related Questions
- `int (dt)/(t^2 sqrt(t^2 - 16))` Evaluate the integral
- 2 Educator Answers
- `int t sec^2 2t dt` Evaluate the integral
- 1 Educator Answer
- `int 5^t sin(5^t) dt` Evaluate the indefinite integral.
- 1 Educator Answer
- Evaluate integral from 0 to sqrt3 of (t^2-1)/(t^4-1) dt.
- 1 Educator Answer
- `int_0^(pi/4)(sec^2 (t))dt` Evaluate the integral.
- 1 Educator Answer