The given function is \( \displaystyle \sin^{3}(2x + 1) \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \( \int \sin^{3}(2x + 1) \, dx \).
Solution:
We know that \( \displaystyle \sin^{3}\theta = \frac{3\sin\theta \,- \sin(3\theta)}{4} \).
Let \( \theta = 2x + 1 \), then: \( \displaystyle \sin^{3}(2x + 1) = \frac{3\sin(2x + 1) \,- \sin(6x + 3)}{4} \).
Now compute the integral:
\( \begin{aligned} &\int \sin^{3}(2x + 1) \, dx \\ &= \int \frac{3\sin(2x + 1)\, – \sin(6x + 3)}{4} \, dx \\ &= \frac{1}{4} \left[ 3 \int \sin(2x + 1) \, dx \,- \int \sin(6x + 3) \, dx \right] \cdots (1)\\ \end{aligned} \)For the first integral, let \(u = 2x + 1\), then \(du = 2 \, dx\):
\( \begin{aligned} &\int \sin(2x + 1) \, dx \\ &= \frac{1}{2} \int \sin(u) \, du \\ &= -\frac{1}{2} \cos(u) + C1, \quad \text{C1 is a constant} \\ &= -\frac{1}{2} \cos(2x + 1) + C1 \\ \end{aligned} \)For the second integral, let \(v = 6x + 3\), then \(dv = 6 \, dx\):
\( \begin{aligned} &\int \sin(6x + 3) \, dx \\ &= \frac{1}{6} \int \sin(v) \, dv \\ &= -\frac{1}{6} \cos(v) + C2, \quad \text{C2 is a constant} \\ &= -\frac{1}{6} \cos(6x + 3) + C2 \\ \end{aligned} \)Substitute the first and second integrals back in (1):
\( \begin{aligned} &\int \sin^{3}(2x + 1) \, dx \\ &= \frac{1}{4} \left[ 3 \left( -\frac{1}{2} \cos(2x + 1) \right) – \left( -\frac{1}{6} \cos(6x + 3) \right) \right] + C, \quad C = C1+C2, \text{C is a constant} \\ &= \frac{1}{4} \left[ -\frac{3}{2} \cos(2x + 1) + \frac{1}{6} \cos(6x + 3) \right] + C \\ &= -\frac{3}{8} \cos(2x + 1) + \frac{1}{24} \cos(6x + 3) + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin^{3}(2x + 1) \, dx = -\frac{3}{8} \cos(2x + 1) + \frac{1}{24} \cos(6x + 3) + C } \\ \)Please let me know in the comments if you find any errors in this solution.