Answer:
[tex]-\frac{1}{4} sin(x)+\frac{1}{6} sin(3x)-\frac{1}{20} sin(5x)+C[/tex]
Step-by-step explanation:
We begin with the integral [tex]\int{sin^2(x)cos(3x)} \, dx[/tex]
First, we can apply the power reducing formula to [tex]sin^2(x)[/tex]
This formula states: [tex]sin^2(x)=\frac{1}{2} -\frac{1}{2} cos(2x)[/tex]
This gives us
[tex]\int{(\frac{1}{2} -\frac{1}{2} cos(2x))(cos(3x)} \, dx \\\\\int{(\frac{1}{2}cos(3x) -(\frac{1}{2} cos(2x)cos(3x)} \, dx \\\\\frac{1}{2} \int{cos(3x)} \, dx -\frac{1}{2} \int{cos(2x)cos(3x)} \, dx[/tex]
Now, we can use integrate the first integral
[tex]\frac{1}{2} \int{cos(3x)} \, dx\\u=3x\\du=3dx\\\\\frac{1}{6} \int{3cos(u)} \, du\\\\\frac{1}{6} sin(u)+C\\\\\frac{1}{6} sin(3x)+C[/tex]
And now we can begin to integrate the second
[tex]-\frac{1}{2} \int{cos(2x)cos(3x)} \, dx[/tex]
To integrate this, we need to use the Product-to-sum formula, which states
[tex]cos(\alpha )cos(\beta )=\frac{1}{2} [cos(\alpha +\beta )+cos(\alpha -\beta )[/tex] . For this formula, we will use [tex]\alpha =3x\\\beta =2x[/tex]
This gives us
[tex]-\frac{1}{2} \int{\frac{1}{2}[cos(5x)+cos(x)] } \, dx \\\\-\frac{1}{4} \int{[cos(5x)+cos(x)] } \, dx\\\\-\frac{1}{4}\int{cos(5x)} \, dx -\frac{1}{4}\int{cos(x)} \, dx[/tex]
We can then use the same process of u-substitution as the previous to get the answer of [tex]-\frac{1}{20} sin(5x)-\frac{1}{4} sin(x)+C[/tex]
Lastly, we can add the values of the two integrals together to give us the final solution of
[tex]-\frac{1}{4} sin(x)+\frac{1}{6} sin(3x)-\frac{1}{20} sin(5x)+C[/tex]
