Assuming you mean natural logs (as 'log' by itself means logs to any base):
∫ln x dx = x ln x - x + C
let y = ln x → ey = x → ey dy = dy
→ ∫ln x dx = ∫y ey dy
But ∫uv = u∫v - ∫u'∫v
let u = y → u' = 1
let v = ey
→ ∫y ey dy = y∫ey dy - ∫∫ ey dy
= y ey - ey + C
= (ln x) x - x + C
= x ln x - x + C
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
integral of e to the power -x is -e to the power -x
The integral of x cos(x) dx is cos(x) + x sin(x) + C
The integral of 2-x = 2x - (1/2)x2 + C.
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
if you mean e to the x power times log of x, it is e to the x divided by x
The integral of 3x is ln(3)*3x. Take the natural log of the base and multiply it by the base raised to the power.
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
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
integral (a^x) dx = (a^x) / ln(a)
the integral of ln(sin(x)) is: -x*ln|1 - e2ix| + x*ln|sin(x)| + (i/2)*(x2 + Li2(e2ix)) + C where Li2 is the second order ploylogarithmic function.
it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.
The Web site integrals.wolfram.com gives the following:integral of sin2x/x = (1/2) (log x - Ci(2 x))Ci is the cosine integral, a special function. Look at the site for a more detailed description.What this really means is that this integral can NOT be solved with the so-called elementary functions, i.e., using only polynomials, roots, trigonometric functions, natural logarithms, and the inverses of some of these.The Web site integrals.wolfram.com gives the following:integral of sin2x/x = (1/2) (log x - Ci(2 x))Ci is the cosine integral, a special function. Look at the site for a more detailed description.What this really means is that this integral can NOT be solved with the so-called elementary functions, i.e., using only polynomials, roots, trigonometric functions, natural logarithms, and the inverses of some of these.The Web site integrals.wolfram.com gives the following:integral of sin2x/x = (1/2) (log x - Ci(2 x))Ci is the cosine integral, a special function. Look at the site for a more detailed description.What this really means is that this integral can NOT be solved with the so-called elementary functions, i.e., using only polynomials, roots, trigonometric functions, natural logarithms, and the inverses of some of these.The Web site integrals.wolfram.com gives the following:integral of sin2x/x = (1/2) (log x - Ci(2 x))Ci is the cosine integral, a special function. Look at the site for a more detailed description.What this really means is that this integral can NOT be solved with the so-called elementary functions, i.e., using only polynomials, roots, trigonometric functions, natural logarithms, and the inverses of some of these.
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
integral of e to the power -x is -e to the power -x
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)