what is the integral of sin^3xdx???

Answer 1

Integrate sin³x with respect to x.

First do some algebraic manipulation to make it easier to integrate.

sin³x = sinx(sin²x) = sinx(1 - cos²x) = sinx - (sinx)(cos²x)

Now we can integrate.

∫sin³x = ∫[sinx - (sinx)(cos²x)]dx

= -cosx + (1/3)cos³x + C

Answer 2

i[..] = integral sin^3(x) dx=

i[sin^2(x).sin(x) dx]=

i[ (1-cos^2(x) ).sin(x) dx] => u=cos(x) => du= -sin(x) dx => sin(x) dx= - du

=> i[..] = i[ (1-u^2)(-du) ]

i[..] = -(u-(u^3)/3) = (u^3)/3 - u + C. Now substitute u= cos(x)

i[..] = ([cos^3(x)]/3) - cos(x) + C
= 1/3(cos^3(x)) - cos(x) + C

Answer 3

-1/3cosx(2 + sin^2(x)) + C

If you differentiate this (carefully!!), you'll get your integrand

Answer 4

(1/4)sec(x)sin^4(x)