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- Improper integral
- Double Integral
- Triple Integral
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How to use it?
- Integral of d{x}:
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Integral of 1/(x²-1)
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Integral of x^2/(1+x^2)^2
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Integral of e^x/(e^x+2)
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Integral of e^(x^2)*x^3
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Identical expressions
- du/cosu
- du divide by co sinus of e of u
- du divide by cosu
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Similar expressions
- du/cos(u)
Integral of du/cosu dx
The solution
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[src]
0 / | | 1 | ------ du | cos(u) | / 0
$$\int\limits_{0}^{0} \frac{1}{\cos{\left(u \right)}}\, du$$
Integral(1/cos(u), (u, 0, 0))
Detail solution
We have the integral:
/ | | 1 | ------ du | cos(u) | /
The integrand
1 ------ cos(u)
Multiply numerator and denominator by
cos(u)
we get
1 cos(u)
------ = -------
cos(u) 2
cos (u)Because
sin(a)^2 + cos(a)^2 = 1
then
2 2 cos (u) = 1 - sin (u)
transform the denominator
cos(u) cos(u) ------- = ----------- 2 2 cos (u) 1 - sin (u)
do replacement
u = sin(u)
then the integral
/ | | cos(u) | ----------- du | 2 = | 1 - sin (u) | /
/ | | cos(u) | ----------- du | 2 = | 1 - sin (u) | /
Because du = du*cos(u)
/ | | 1 | ------ du | 2 | 1 - u | /
Rewrite the integrand
1 1
----- + -----
1 1 - u 1 + u
------ = -------------
2 2
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(u)
The answer
/ | | 1 log(1 + sin(u)) log(-1 + sin(u)) | ------ du = --------------- - ---------------- + C0 | cos(u) 2 2 | /
where C0 is constant, independent of u
The answer (Indefinite)
[src]
/ | | 1 log(1 + sin(u)) log(-1 + sin(u)) | ------ du = C + --------------- - ---------------- | cos(u) 2 2 | /
$$\int \frac{1}{\cos{\left(u \right)}}\, du = C - \frac{\log{\left(\sin{\left(u \right)} - 1 \right)}}{2} + \frac{\log{\left(\sin{\left(u \right)} + 1 \right)}}{2}$$
The graph
Use the examples entering the upper and lower limits of integration.