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- Improper integral
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How to use it?
- Integral of d{x}:
- Integral of sin^-1(x)
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Integral of 1/x(x+1)
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Integral of x(x^2+1)^3
- Integral of e^(1/x)
- Derivative of:
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sin^-1(x)
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Identical expressions
- sin^- one (x)
- sinus of to the power of minus 1(x)
- sinus of to the power of minus one (x)
- sin-1(x)
- sin-1x
- sin^-1x
- sin^-1(x)dx
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Similar expressions
- sin^-1x
- sin^+1(x)
Integral of sin^-1(x) 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 | 1*------ dx | sin(x) | /
The integrand
1 ------ sin(x)
Multiply numerator and denominator by
sin(x)
we get
1 1*sin(1*x)
------ = ----------
sin(x) 2
sin (1*x) Because
sin(a)^2 + cos(a)^2 = 1
then
2 2 sin (x) = 1 - cos (x)
transform the denominator
1*sin(1*x) 1*sin(1*x) ---------- = ------------- 2 2 sin (1*x) 1 - cos (1*x)
do replacement
u = cos(x)
then the integral
/ | | 1*sin(1*x) | ------------- dx | 2 = | 1 - cos (1*x) | /
/ | | 1*sin(1*x) | ------------- dx | 2 = | 1 - cos (1*x) | /
Because du = -dx*sin(x)
/ | | 1*sin(1*x) | ------------- du | 2 | 1 - cos (1*x) | /
Rewrite the integrand
1*sin(1*x) 1*-1 / 1 1 \
------------- = ----*|----- + -----|
2 2 \1 - u 1 + u/
1 - cos (1*x) then
/ /
| |
| 1 | 1
| ----- du | ----- du
/ | 1 + u | 1 - u
| | |
| 1*sin(1*x) / / =
| ------------- du = - ----------- - -----------
| 2 2 2
| 1 - cos (1*x)
|
/
= u*sin(x)/(1 - cos(x)^2)
do backward replacement
u = cos(x)
The answer
/ | | 1 log(-1 + cos(x)) log(1 + cos(x)) | 1*------ dx = ---------------- - --------------- + C0 | sin(x) 2 2 | /
where C0 is constant, independent of x
The answer (Indefinite)
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/ | | 1 log(-1 + cos(x)) log(1 + cos(x)) | ------ dx = C + ---------------- - --------------- | sin(x) 2 2 | /
$${{\log \left(\cos x-1\right)}\over{2}}-{{\log \left(\cos x+1\right)
}\over{2}}$$
Use the examples entering the upper and lower limits of integration.