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
int cos3x=sin3x/3+c
x integration 0 x integration x siny/ydydx
Integration for inverse tangent of square 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)
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)
-cos x + C
(2/3)x^(3/2)
-x-1
The integral of f'(x) = 1 is f(x) = x + c