= 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
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
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
.5(x-sin(x)cos(x))+c
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
-cos(x) + constant
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.
The integral of cosine cubed is sinx- 1/3 sin cubed x + 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
-cos x + Constant
The integral of x cos(x) dx is cos(x) + x sin(x) + C
sin2x + c
(1/8)(x-sin 4x)
- cos(1 - X) + C
∫ cos(x) dx = -sin(x) + C
.5(x-sin(x)cos(x))+c
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
∫ cot(x) dx is written as: ∫ cos(x) / sin(x) dx Let u = sin(x). Then, du = cos(x) dx, giving us: ∫ 1/u du So the integral of 1/u is ln|u|. So the answer is ln|sin(x)| + c