Q: Integration of log of sin x?

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

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)

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.

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

(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True

Related questions

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

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.

âˆ« sin(x) dx = -cos(x) + CC is the constant of integration.

âˆ« cos(x) dx = sin(x) + CC is the constant of integration.

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)

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 + âˆš(1-x2)] So in this case: = -i*log[i*6 + âˆš(1-36)] = -i*log[6*i + âˆš(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i

âˆ« cot(x) dx = ln(sin(x)) + CC is the constant of integration.

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)

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)

âˆ« sin(x)/cos2(x) dx = sec(x) + C C is the constant of integration.

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.

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