How do you find the integral of $sin^3[x]dx$ ?

Question

248439 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-19T18:44:23

$intsin^3(x)dx = 1/3cos^3(x)-cos(x)+C$

$intsin^3(x)dx = intsin(x)(1-cos^2(x))dx$

$=intsin(x)dx - intsin(x)cos^2(x)dx$

For the first integral:

$intsin(x)dx = -cos(x)+C$

For the second integral, using substitution:

Let $u = cos(x) => du = -sin(x)dx$
Then

$-intsin(x)cos^2(x)dx = intu^2du$

$=u^3/3+C$

$=1/3cos^3(x)+C$

Putting it all together, we get our final result:

$intsin^3(x)dx = intsin(x)dx-intsin(x)cos^2(x)dx$

$=-cos(x)+1/3cos^3(x)+C$

How do you find the integral of sin^3[x]dx sin^3[x]dx?