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- Improper integral
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How to use it?
- Integral of d{x}:
-
Integral of (-1)^x
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Integral of xln(2-x)
- Integral of xsin^4(x^2)cos^4(x^2)
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Integral of x*dx/e^(3*x^2+4)
- Graphing y =:
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1/sin(2*x)
- Derivative of:
- 1/sin(2*x)
- Limit of the function:
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1/sin(2*x)
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Identical expressions
- one /sin(two *x)
- 1 divide by sinus of (2 multiply by x)
- one divide by sinus of (two multiply by x)
- 1/sin(2x)
- 1/sin2x
- 1 divide by sin(2*x)
- 1/sin(2*x)dx
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Similar expressions
- ctgx+1/sin^2(x)
- 1/(sin2x+2sinx)
- (1-1/sin^2x)
- 1/(sin(2*x)*sin(x))
- (sinx+1/sin^2x)
Integral of 1/sin(2*x) dx
The solution
You have entered
[src]
1 / | | 1 | -------- dx | sin(2*x) | / 0
$$\int\limits_{0}^{1} \frac{1}{\sin{\left(2 x \right)}}\, dx$$
Integral(1/sin(2*x), (x, 0, 1))
Detail solution
We have the integral:
/ | | 1 | -------- dx | sin(2*x) | /
The integrand
1 -------- sin(2*x)
Multiply numerator and denominator by
sin(2*x)
we get
1 sin(2*x)
-------- = ---------
sin(2*x) 2
sin (2*x)Because
sin(a)^2 + cos(a)^2 = 1
then
2 2 sin (2*x) = 1 - cos (2*x)
transform the denominator
sin(2*x) sin(2*x) --------- = ------------- 2 2 sin (2*x) 1 - cos (2*x)
do replacement
u = cos(2*x)
then the integral
/ | | sin(2*x) | ------------- dx | 2 = | 1 - cos (2*x) | /
/ | | sin(2*x) | ------------- dx | 2 = | 1 - cos (2*x) | /
Because du = -2*dx*sin(2*x)
/ | | -1 | ---------- du | / 2\ | 2*\1 - u / | /
Rewrite the integrand
-1 -1 / 1 1 \ ---------- = ---*|----- + -----| / 2\ 2*2 \1 - u 1 + u/ 2*\1 - u /
then
/ /
| |
| 1 | 1
| ----- du | ----- du
/ | 1 + u | 1 - u
| | |
| -1 / / =
| ---------- du = - ----------- - -----------
| / 2\ 4 4
| 2*\1 - u /
|
/
= -log(1 + u)/4 + log(-1 + u)/4
do backward replacement
u = cos(2*x)
The answer
/ | | 1 log(1 + cos(2*x)) log(-1 + cos(2*x)) | -------- dx = - ----------------- + ------------------ + C0 | sin(2*x) 4 4 | /
where C0 is constant, independent of x
The answer (Indefinite)
[src]
/ | | 1 log(1 + cos(2*x)) log(-1 + cos(2*x)) | -------- dx = C - ----------------- + ------------------ | sin(2*x) 4 4 | /
$$\int \frac{1}{\sin{\left(2 x \right)}}\, dx = C + \frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{4} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{4}$$
The graph
The answer
[src]
pi*I
oo + ----
4
$$\infty + \frac{i \pi}{4}$$
=
=
pi*I
oo + ----
4
$$\infty + \frac{i \pi}{4}$$
oo + pi*i/4
The graph
Use the examples entering the upper and lower limits of integration.