0

Integration of log of sin x?

Updated: 12/20/2022

Wiki User

14y ago

the integral of ln(sin(x)) is:

-x*ln|1 - e2ix| + x*ln|sin(x)| + (i/2)*(x2 + Li2(e2ix)) + C

where Li2 is the second order ploylogarithmic function.

Wiki User

14y ago

Earn +20 pts
Q: Integration of log of sin x?
Submit
Still have questions?
Related questions

Integration of sin x?

Integration by Parts is a special method of integration that is often useful when two functions.

Find the derivative y equals 10 to the sin3x power?

Y=10^sin(x) The derivative is: (log(5)+log(2))*cos(x)*2^sin(x)*5^sin(x) Use the chain rule, product rule, and power rules combined with sin(x) rule.

What is the integral of the sine of x with respect to x?

&acirc;&circ;&laquo; sin(x) dx = -cos(x) + CC is the constant of integration.

What is the integral of the cosine of x with respect to x?

&acirc;&circ;&laquo; cos(x) dx = sin(x) + CC is the constant of integration.

Integral of cos x cot x?

cos(x) cot(x) = cos(x) * 1/(tan(x)) = cos(x) * 1 / (sinx(x) / cos(x)) = cos2(x) / sin(x) = (1-sin2(x)) / sin(x) = 1/sin(x) - sin(x) so the antiderivative of cos(x)cot(x) = log[abs(tan(x/2))]+cos(x) This can also be written as log[abs((sin(x)/(cos(x)+1))]+cos(x) if we want everything in terms of x and not (x/2). The two answers are, of course, the same. where log(x) refers to the natural log, often written ln(x). We might write ln[|sin(x)|/|cos(x)+1|] +cos(x)

How do you solve sin x equals 6?

First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + &acirc;&circ;&scaron;(1-x2)] So in this case: = -i*log[i*6 + &acirc;&circ;&scaron;(1-36)] = -i*log[6*i + &acirc;&circ;&scaron;(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i

What is the integral of the cotangent of x with respect to x?

&acirc;&circ;&laquo; cot(x) dx = ln(sin(x)) + CC is the constant of integration.

How you can defferentiate an integral?

If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)

How do you differentiate 9 to the power sin squared x sorry I wasn't allowed to use mathematical notation when asking?

derivative of 9[sin(x)]^2 is found by first letting u(x)=[sin(x)]^2. Note that sin2x = [sin(x)]^2, and the ^2 means raising the base to the exponent 2. Find the d(9u(x))/dx using the chain rule. d( 9u(x) )/dx = (d(9u)/du)(du/dx ) , by the chain rule. So we need: d(9u)/du = 9ulog(9) du/dx = d( [sin(x)]^2 )/dx = 2sin(x) d( sin(x) )/dx = 2sin(x)cos(x) Puttin this together gives: d( 9u(x) )/dx = 9u(log(9)) 2sin(x)cos(x) Now substitute in u(x) = [sin(x)]^2. d( 9u(x) )/dx = 9[sin(x)]^2(log(9)) 2sin(x)cos(x) = 2 log(9) 9[sin(x)]^2sin(x)cos(x) or = log(9) 9[sin(x)]^2sin(2x)

What is the integral of the sine of x divided by the cosine squared of x with respect to x?

&acirc;&circ;&laquo; sin(x)/cos2(x) dx = sec(x) + C C is the constant of integration.

How do you prove that the derivative of sec x is equal to sec x tan x?

Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.

An equation that has variables in both the numerator and denominator?

It has no specific name. For example f(x) = sin(x)/log(x) where x not equal to 1