0
= 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
Still curious? Ask our experts.
Chat with our AI personalities
Judy
Taiga
LaoInt(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
Add your answer:
Earn +20 pts
Integral of 1 divided by sinx cosx?
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 of sine squared x?
.5(x-sin(x)cos(x))+c
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
What is integral square root of sin x.dx?
-cos(x) + constant
What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
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.
