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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)

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Q: Integral of cos x cot x?

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Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x

sin integral is -cos This is so because the derivative of cos x = -sin x

The integral of x cos(x) dx is cos(x) + x sin(x) + C

sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)

Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.

== cot(x)== 1/tan(x) = cos(x)/sin(x) Now substitute cos(x)/sin(x) into the expression, in place of cot(x) So now: sin(x) cot(x) cos(x) = sin(x) cos(x) (cos(x)/sin(x) ) sin(x) cos(x) cos(x)/sin(x) The two sin(x) cancel, leaving you with cos(x) cos(x) Which is the same as cos2(x) So: sin(x) cot(x) cos(x) = cos2(x) ===

Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1

∫ cot(x) dx is written as: ∫ cos(x) / sin(x) dx Let u = sin(x). Then, du = cos(x) dx, giving us: ∫ 1/u du So the integral of 1/u is ln|u|. So the answer is ln|sin(x)| + c

The trig identaty of cot(x) is cos(x)/sin(x) so then if we want to evaluate cot (68) deg. we just plug into the identady. so cos(68)/sin(68)=.404

cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)

-cos x + Constant

(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.

Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)

sin2x + c

the questions is 2x=(cot^2 x-1)/(cot^2 x+1)

y = sec(x)*cot(x)*cos(x)To solve this trigonometric equation, you need to know these identities:sec(x) = 1/(cos(x))cot(x) = 1/(tan(x)) = (cos(x))/(sin(x))Now substitute these identities into the original equation:y = (1/cos(x))*((cos(x))/(sin(x)))*cos(x)Now cancel out the terms that are similar in the numerator and denominator to leave you with:y = (1/(sin(x)))*cos(x)y = (cos(x))/(sin(x))From the aforementioned known identity, the final simplified trigonometric equation becomes:y = cot(x)

The TI-83 does not have the cot button, however, if you type 1/tan( then this will work the same as the cot since cot=1/tan. The other way to do this is to type (cos(x))/(sin(x)) where x is the angle you're looking for. This works because cot=cos/sin

2 cot(x) + 1 = -1 2 cot(x) = -2 cot(x) = -1 cos(x)/sin(x) = -1 cos(x) = - sin(x) x = 135°, 315°, 495°, ... another one every 180 degrees

The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity

The integral of cot(x)dx is ln|sin(x)| + C

I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3

- cos(1 - X) + C

cot[x]= -1 cot[x] = cos[x] / sin[x] cos[x] / sin[x] = -1 cos[x] = -sin[x] |cos[x]| = |sin[x]| at every multiple of Pi/4 + Pi/2. However, the signs disagree at 3Pi/4 + nPi, where n is an integer.

convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx

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