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int cos3x=sin3x/3+c

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Q: Integration of cos3 x
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Evaluate Limit x tends to 0 x minus sinx divided by tanx minus x?

lim (x→0) [(x - sin x)/(tan x - x)]Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equalslim (x→0) [(x - sin x)'/(tan x - x)']= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)= lim (x→0) [(sin x)/(2sin x/cos3 x)]= lim (x→0) [(sin x cos3 x)/2sin x]= lim (x→0) (cos3 x/2)= 1/2Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.


Integration of sinx is?

-cos x + C


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S 2/x d/x bring the constant 2 out in front of the sign of integration 2 S 1/x dx you should know the integration of 1/x 2*ln(x) + C


What is the integral of ln2?

The integral of ln(2) is a constant multiple of x times the natural logarithm of 2, plus a constant of integration. In other words, the integral of ln(2) with respect to x is x * ln(2) + C, where C is the constant of integration. This integral represents the area under the curve of the natural logarithm of 2 function with respect to x.


What is integration of x to the power x with respect to x?

Apparently that can't be solved with a finite number of so-called "elementary functions". You can get the beginning of the series expansion here: http://www.wolframalpha.com/input/?i=integrate+x^x

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Evaluate Limit x tends to 0 x minus sinx divided by tanx minus x?

lim (x→0) [(x - sin x)/(tan x - x)]Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equalslim (x→0) [(x - sin x)'/(tan x - x)']= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)= lim (x→0) [(sin x)/(2sin x/cos3 x)]= lim (x→0) [(sin x cos3 x)/2sin x]= lim (x→0) (cos3 x/2)= 1/2Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.


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Integration of sinx is?

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