Take f(x) = cos(3x)
∫ f(x) dx
= ∫ cos(3x) dx
Take u=3x → du = 3dx
= ∫ 1/3*cos(u) du
= 1/3*∫ cos(u) du
= 1/3*sin(u) + C, C ∈ ℝ
= 1/3*sin(3x) + C
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
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)
Integration for inverse tangent of square x
It is cosh(x) + c where c is a constant of integration.
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 by Parts is a special method of integration that is often useful when two functions.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
_____ 1 x (x+4) (x-3)
(2/3)x^(3/2)
The integral of sec(x) with respect to x is ln|sec(x) + tan(x)| + C, where C is the constant of integration. This result can be derived using integration techniques such as substitution or integration by parts. The integral of sec(x) is a common integral in calculus and is often used in trigonometric integrals.
The integration formulas covered in the second PUC syllabus primarily include basic integration techniques such as integration of power functions, trigonometric functions, exponential functions, and logarithmic functions. Key formulas include ∫ x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1, ∫ sin(x) dx = -cos(x) + C, and ∫ e^x dx = e^x + C. Additionally, students learn about integration by substitution and integration by parts. Understanding these fundamental formulas is essential for solving various problems in calculus.