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