int x ln5x dx by parts u = ln5x du = 1/5x or 5x^-1 dv = x v = 1/2x^2 uv - int v du ln5x 1/2x^2 - int 1/2x^2 5x^-1 1/2ln5x*x^2 - 1/6x^3 5x + C
By using: ∫uv = u∫v - ∫u'∫v twice. ∫x2exdx let u = x2, v = ex, then: ∫x2exdx = x2∫exdx - ∫d/dx(x2)∫exdx dx = x2ex - ∫2xexdx Again, let u = 2x, v = ex, then: = x2ex - (2x∫exdx - ∫d/dx(2x)∫exdx dx) = x2ex - 2xex +∫2exdx = x2ex - 2xex + 2ex + C
The Least Common Multiple (LCM) of x and 2x is the smallest multiple that is divisible by both x and 2x. Since 2x is a multiple of x, the LCM is simply 2x. This is because any multiple of x will also be a multiple of 2x, making 2x the smallest common multiple of x and 2x.
If 2x + 5 = 11, then 11 - 5 = 2x 6 = 2x 3 = x
sin(2x), cos(2x), cosec(2x), sec(2x), tan(x) and cot(x) are all possible functions.
2x+x is 3x
(2x + 1) + (x*x - 2x + 1) = x^2 + 2x - 2x + 1 + 1 = x^2 + 2
2X * 2X * 2X = 8X^3 2X^3 is saying 2*( X * X * X) = 2X^3
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
2x + 3 = x 2x + 3 - 3 = x - 3 2x = x - 3 2x - x = x - 3 - x x = -3
8x4
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
4
2x - x = x
int x ln5x dx by parts u = ln5x du = 1/5x or 5x^-1 dv = x v = 1/2x^2 uv - int v du ln5x 1/2x^2 - int 1/2x^2 5x^-1 1/2ln5x*x^2 - 1/6x^3 5x + C
By using: ∫uv = u∫v - ∫u'∫v twice. ∫x2exdx let u = x2, v = ex, then: ∫x2exdx = x2∫exdx - ∫d/dx(x2)∫exdx dx = x2ex - ∫2xexdx Again, let u = 2x, v = ex, then: = x2ex - (2x∫exdx - ∫d/dx(2x)∫exdx dx) = x2ex - 2xex +∫2exdx = x2ex - 2xex + 2ex + C