To integrate such a function, you must use the u-substitution technique. Let u = 4x; therefore du = 4 dx. Then du/4 = dx. The integral becomes : Int(cos 4x dx) = Int[(cos u)(du/4)] = ¼ Int(cos u du) = ¼ (sin u) + C = ¼ sin 4x + C
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Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
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Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
d/dx (1-x)=d/dx (1)-d/dx(x)=0-1=-1
The product rule states: d/dx uv = vdu/dx + udv/dxIt is not quite clear what your denominator is:"over 1 + (x squared)": d/dx (x/1+x2)d/dx(x/1+x2) = d/dx x(1 + x2)-1 = (1 + x2)-1 d/dx x + x d/dx (1 + x2)-1= (1 + x2)-1 + x -2x (1 + x2)-2= (1 + x2)-2 (1 + x2 - 2x2)= (1 - x2) / (1 + x2)2"over (1 + x) [all] squared": d/dx (x/(1+x)2)d/dx(x/(1+x)2) = d/dx x(1 + x)-2 = (1 + x)-2 d/dx x + x d/dx (1 + x)-2= (1 + x)-2 + x -2 (1 + x)-3= (1 + x)-3 (1 + x - 2x)= (1 - x)/(1 + x)3If you prefer, you can use the quotient rule: d/dx (u/v) = (v du/dx - udv/dx) / v2