what is the integral of sinx(1-(sinx)^2)?

Answer 1

1-(sinx)^2 =(cosx)^2

so the integral is (cosx)^2 sinx dx

if you write cosx=u then sinx dx =-du

and the integral -u^2 du

which is -u^3/3 so the result is (-1/3) (cosx)^3 +C

Answer 2

Integral of sinx(1-(sinx)^2) is found by using the substitution method.
Sinx(1-(sinx)^2) is simplified to sinx((cosx)^2)
by the identity (sinx)^2 + (cosx)^2 =1 , whereby 1 - (sinx)^2 = (cosx)^2
Now the integral becomes (cosx)^2 * sinx
Substituting cosx = t, we have sinx* dx = dt
Now the integral becomes (t^2) dt which is equal to t^3/3
Hence the complete integral is ((cosx)^3) / 3 + C

Answer 3

Notice that (1-(sin x)^2) = (cos x)^2 (Pythagorean identity)

So: int[(sin x)(1-(sin x)^2)] = INT [ sin x * (cos x)^2]

let u=cos x --> du = -sin x dx

= INT -[u^2]dx = -(1/3) u^3 + C === -(1/3) (cos x)^3 + C

Answer 4

sin(x)(cos²x)=
-1/3cos^3(x)