.5(x-sin(x)cos(x))+c
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
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
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∫sin²x cos²x dx = ∫(1-cos²x)cos²x dx =∫cos²xdx-⌠cos²xcos²xdx =1/2⌠1+cos2x dx-1/2⌠[(1+cos2x)(1+cos2x)] Do the operations, distributions, arrange common numbers, and try to sort out the factors as a polynomiom. Then, =1/2x+1/4sin2x-1/2x-1/2sin2x-1/4x-1/16sin4x =-1/4x-1/16sin4x
.5(x-sin(x)cos(x))+c
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
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
sin integral is -cos This is so because the derivative of cos x = -sin x
The integral of x cos(x) dx is cos(x) + x sin(x) + C
Sin squared, cos squared...you removed the x in the equation.
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tan(x) + C d/dx tan(x) = d/dx (sin(x))/(cos(x)) = (sin^2(x)+cos^2(x))/(cos^2(x)) = 1/(cos^2(x)) = sec^2(x) NEVER FORGET THE CONSTANT!
∫ cos(x)/sin2(x) dx = -cosec(x) + C C is the constant of integration.
2 x cosine squared x -1 which also equals cos (2x)
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