How do you find the integral of sin3[x]dx?

∫sin3(x)dx

Solution

∫sin3(x)dx=∫sin(x)(1−cos2(x))dx

=∫sin(x)dx−∫sin(x)cos2(x)dx

Now let us consider the first integral

We know that

∫sin(x)dx=−cos(x)+C

Now for the second integral,

We will use the substitution

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

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