1/3ln(sin3x) + C
8
Do you mean the Convolution Integral?
The integral of -x2 is -1/3 x3 .
2 2x makes no sense. If you meant the integral of 2x, it is x2 + C. If you meant the integral of 4x, it is 2x2 + C. If you meant the integral of 2x2, it is 2/3 x3 + C.
ln(sinx) + 1/3ln(sin3x) + 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 start, try breaking down sin3x using a double angle formula. Message me if you need more help!
1/3ln(sin3x) + C
int cos3x=sin3x/3+c
sin(3x) = 3sin(x) - 4sin^3(x)
According to de Moivre's formula, cos3x + isin3x = (cosx + isinx)3 = cos3x + 3cos2x*isinx + 3cosx*i2sin2x + i3sin3x Comparing the imaginary parts, isin3x = 3cos2x*isinx + i3sin3x so that sin3x = 3cos2x*sinx - sin3x = 3*(1-sin2x)sinx - sin3x = 3sinx - 4sin3x
This has no solution. The sine of any number is always between 1 and -1.
That's 1/2 .(You have to use l'Hospital's rule.)
8
Integral in Tagalog: mahalaga
In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.