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)] dx
since this is multiplicative, switch it around
int[cos(15X)cos(10X)] dx
int[cos(15X - 10X)/2(15 -10) + cos(15X + 10X)/2(15 + 10)] dx
int[cos(5X)/10 + cos(25X)/50] dx
= 1/10sin(5X) + 1/50sin(25X) + C
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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
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
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
∫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