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Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
What is 32-5x equals 3x plus 8?
32-5x=3x+8 32-5x-8=3x+8-8 24-5x=3x 24-5x+5x=3x+5x 24=8x x=24/8=3
What is 5x-1 7 3x?
what is |5x-1|+7=3x
5x plus 5 equals 8 - 3x?
5x + 5 = 8 -3x 5x = 3 -3x 8x = 3 x = 0.375
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
What is the integral of sin 5X cos 10X dX?
use product-to-sumformula sin u cos v =1/2 [sin(u+v)+ sin(u-v)]so you get1/2 int: (sin 15x) - (sin5x) dxsplit it1/2 int: sin 15x dx- 1/2 int: sin 5x dxusing substitution you can conclude that1/30 int: sin u du- 1/10 in sin w dw(you get the fraction change when you set dx=duand dw)so then- (cos u)/30 + (cos w)/10replace the substitution(cos 5x)/10 - (cos 15x)/30 + Constant
Integration of cos 5x?
The integral of cos 5x is 1/5 sin (5x)
What is the answer to this problem cos 5x plus 49 sin 3x plus 57?
There are operators missing. The only question that I can see to make sense is: Solve x for: cos(5x + 49°) = sin(3x + 57°) It's been a while since I did this kind of problem, so there may be more solutions to the ones I give here: Cosθ = sin(90 - θ) → cos(5x + 49°) = sin(3x + 57°) → sin(90° - (5x + 49°)) = sin(3x + 57°) → sin(41° - 5x) = sin(3x + 57°) Thus: 41° - 5x = 3x + 57° → 8x = -16° → x = -2° But as sin and cos are cyclic with a period of 360°, -2° = 360° - 2= 358° → x = 358° + 360°n where n = 0, 1, 2, .... But sin θ = sin(180° - θ) which means that 180 - (41° - 5x) = 3x + 57° → 5x + 139° = 3x + 57° → 2x = -82° → x = -41° → x = 319° + 360°n where n = 0, 1, 2, 3,... is also a solution set. Thus the solutions are: x = 358° + 360°n x = 319° + 360°n where n = 0, 1, 2, 3, ...
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
5x equals 3x plus 13 fall between what two integers?
Answer: Between integers 6 and 7Math:5x = 3x + 135x -3x = 3x -3x + 132x = 13(2x) : 2 = (13) : 2Int (6) < (x = 6.5) < Int (7)
Int cos 10x cos 15x dx?
There is some kind of formula here, half angle, or some such that I forget, but I do remember the algorithm. So...,int[cos(10X)cos(15X)] dxsince this is multiplicative, switch it aroundint[cos(15X)cos(10X)] dxint[cos(15X - 10X)/2(15 -10) + cos(15X + 10X)/2(15 + 10)] dxint[cos(5X)/10 + cos(25X)/50] dx= 1/10sin(5X) + 1/50sin(25X) + C=========================
How do you solve 5x-2 3x-6?
5x-2 = 3x-6 5x = 3x - 6 +2 5x = 3x - 4 5x - 3x = -4 2x = -4 x = -2
What is 32-5x equals 3x plus 8?
32-5x=3x+8 32-5x-8=3x+8-8 24-5x=3x 24-5x+5x=3x+5x 24=8x x=24/8=3
What is 5x-1 7 3x?
what is |5x-1|+7=3x
6-3x equals 5x-10x plus 10?
6-3x = 5x-10x+10 6-3x = -5x+10 -3x+5x = 10-6 2x = 4 x = 2
What is 3x minus 5x plus x?
3x - 5x + x = -x
