Integrate sin (3x) dx

We have to integrate ∫sin3(x)dx

Solution

∫sin3(x)dx=∫sin(x)(1−cos2(x))dx

=∫sin(x)dx−∫sin(x)cos2(x)dx

Let us consider the first integral

We know that

∫sin(x)dx=−cos(x)+C

Now for the second integral,

Let us assume that cos(x)=u

Hence, du=−sin(x)dx
Therefore

−∫sin(x)cos2(x)dx=∫u2du

=u3/3+C

=1/3cos3(x)+C

Combining all the above obtained values we get

∫sin3(x)dx=∫sin(x)dx−∫sin(x)cos2(x)dx

=−cos(x)+1/3cos3(x)+C

Answer

∫sin3(x)dx=−cos(x)+1/3cos3(x)+C