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- Improper integral
- Double Integral
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How to use it?
- Integral of d{x}:
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Integral of 1/(x+1)^3
- Integral of tan^(-1)x
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Integral of x^3/(x-1)
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Integral of senx
- Derivative of:
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(-1)/cos(x)
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Identical expressions
- (- one)/cos(x)
- ( minus 1) divide by co sinus of e of (x)
- ( minus one) divide by co sinus of e of (x)
- -1/cosx
- (-1) divide by cos(x)
- (-1)/cos(x)dx
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Similar expressions
- 1/x^8-1/cosx+4^x-1/x^2-25
- ((x^2)-1)/cos(x)
- (cosx-1)/(cosx+1)
- (sqrt(tg(x))-1)/((cos(x))^2)
- -((1/cosx)+(1/(3cosx)))
- (1)/cos(x)
- (-1)/cosx
Integral of (-1)/cos(x) dx
The solution
You have entered
[src]
1 / | | -1 | ------ dx | cos(x) | / 0
$$\int\limits_{0}^{1} \left(- \frac{1}{\cos{\left(x \right)}}\right)\, dx$$
Integral(-1/cos(x), (x, 0, 1))
Detail solution
We have the integral:
/ | | -1 | ------ dx | cos(x) | /
The integrand
-1 ------ cos(x)
Multiply numerator and denominator by
cos(x)
we get
-1 -cos(x)
------ = --------
cos(x) 2
cos (x) Because
sin(a)^2 + cos(a)^2 = 1
then
2 2 cos (x) = 1 - sin (x)
transform the denominator
-cos(x) -cos(x) -------- = ----------- 2 2 cos (x) 1 - sin (x)
do replacement
u = sin(x)
then the integral
/ | | -cos(x) | ----------- dx | 2 = | 1 - sin (x) | /
/ | | -cos(x) | ----------- dx | 2 = | 1 - sin (x) | /
Because du = dx*cos(x)
/ | | -1 | ------ du | 2 | 1 - u | /
Rewrite the integrand
-1 -1 / 1 1 \
------ = ---*|----- + -----|
2 2 \1 - u 1 + u/
1 - u then
/ /
| |
| 1 | 1
| ----- du | ----- du
/ | 1 + u | 1 - u
| | |
| -1 / / =
| ------ du = - ----------- - -----------
| 2 2 2
| 1 - u
|
/
= log(-1 + u)/2 - log(1 + u)/2
do backward replacement
u = sin(x)
The answer
/ | | -1 log(-1 + sin(x)) log(1 + sin(x)) | ------ dx = ---------------- - --------------- + C0 | cos(x) 2 2 | /
where C0 is constant, independent of x
The answer (Indefinite)
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/ | | -1 log(-1 + sin(x)) log(1 + sin(x)) | ------ dx = C + ---------------- - --------------- | cos(x) 2 2 | /
$$\int \left(- \frac{1}{\cos{\left(x \right)}}\right)\, dx = C + \frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{2}$$
The graph
The answer
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log(1 - sin(1)) log(1 + sin(1))
--------------- - ---------------
2 2
$$\frac{\log{\left(1 - \sin{\left(1 \right)} \right)}}{2} - \frac{\log{\left(\sin{\left(1 \right)} + 1 \right)}}{2}$$
=
=
log(1 - sin(1)) log(1 + sin(1))
--------------- - ---------------
2 2
$$\frac{\log{\left(1 - \sin{\left(1 \right)} \right)}}{2} - \frac{\log{\left(\sin{\left(1 \right)} + 1 \right)}}{2}$$
log(1 - sin(1))/2 - log(1 + sin(1))/2
The graph
Use the examples entering the upper and lower limits of integration.