How do you find the integral of $\int {sin}^{3} (x) d x$ $int sin^3(x)dx$ ?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-06T02:31:45

$int sin^{3}(x)\ dx=1/3 cos^{3}(x)-cos(x)+C$

Use the identity $sin^{2}(x)=1-cos^{2}(x)$ and then let $u=cos(x)$ so that $du = -sin(x)\ dx$ and

$int sin^{3}(x)\ dx=int(1-cos^{2}(x))sin(x)\ dx=int (u^2-1)\ du$

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

How do you find the integral of int sin^3(x)dx∫sin3(x)dxint sin^3(x)dx?