x integration 0 x integration x siny/ydydx
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
It is cosh(x) + c where c is a constant of integration.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
Integration by Parts is a special method of integration that is often useful when two functions.
int cos3x=sin3x/3+c
x integration 0 x integration x siny/ydydx
Integration for inverse tangent of square x
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
It is cosh(x) + c where c is a constant of integration.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
Integration by Parts is a special method of integration that is often useful when two functions.
_____ 1 x (x+4) (x-3)
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.
(2/3)x^(3/2)
-cos x + C
What is the [int] (x+1)*5^[(x+1)^2]? [int]= integration I got 5^[(x+1)^2]/ 2ln 5. Is this correct?