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- Integral of d{x}:
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Integral of (4x)/(1+x^2)
- Integral of 2x-1/(x-1)(x-2)
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Integral of 2*exp(x)
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Integral of 2/sinx
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Identical expressions
- two /sinx
- 2 divide by sinus of x
- two divide by sinus of x
- 2 divide by sinx
- 2/sinxdx
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Similar expressions
- 2/(sinxcosx)
- (cosx^2)/(sinx^2)
- arcsin^2/((sinx^5))^1/2
- (1-(cos(x))^2)/(sin(x)*cos(x))
- sqrt((sinx^4)+(4*cosx^2))/(sinx^2)
Integral of 2/sinx dx
The solution
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1 / | | 2 | ------ dx | sin(x) | / 0
$$\int\limits_{0}^{1} \frac{2}{\sin{\left(x \right)}}\, dx$$
Integral(2/sin(x), (x, 0, 1))
Detail solution
We have the integral:
/ | | 2 | ------ dx | sin(x) | /
The integrand
2 ------ sin(x)
Multiply numerator and denominator by
sin(x)
we get
2 2*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
2*sin(x) 2*sin(x) -------- = ----------- 2 2 sin (x) 1 - cos (x)
do replacement
u = cos(x)
then the integral
/ | | 2*sin(x) | ----------- dx | 2 = | 1 - cos (x) | /
/ | | 2*sin(x) | ----------- dx | 2 = | 1 - cos (x) | /
Because du = -dx*sin(x)
/ | | -2 | ------ du | 2 | 1 - u | /
Rewrite the integrand
-2 2*(-1) / 1 1 \
------ = ------*|----- + -----|
2 2 \1 - u 1 + u/
1 - u then
/ / / | | | | -2 | 1 | 1 | ------ du = - | ----- du - | ----- du | 2 | 1 + u | 1 - u = | 1 - u | | | / / /
= -log(1 + u) + log(-1 + u)
do backward replacement
u = cos(x)
The answer
/ | | 2 | ------ dx = -log(1 + cos(x)) + log(-1 + cos(x)) + C0 | sin(x) | /
where C0 is constant, independent of x
The answer (Indefinite)
[src]
/ | | 2 | ------ dx = C - log(1 + cos(x)) + log(-1 + cos(x)) | sin(x) | /
$$\int \frac{2}{\sin{\left(x \right)}}\, dx = C + \log{\left(\cos{\left(x \right)} - 1 \right)} - \log{\left(\cos{\left(x \right)} + 1 \right)}$$
The graph
The graph
Use the examples entering the upper and lower limits of integration.