The given function is \( \displaystyle \sin^{2}(2x + 5) \) and we need to compute the integral of this function with respect to \( x \).
Solution:
We know that \(\displaystyle \sin^{2}\theta = \frac{1 – \cos(2\theta)}{2} \).
Let \(\theta = 2x + 5 \). Then, \( \displaystyle \sin^{2}(2x + 5) = \frac{1 – \cos(4x + 10)}{2} \).
Now substitute the value into the integral:
\( \begin{aligned} &\int \sin^{2}(2x + 5) \, dx \\ &= \int \frac{1 – \cos(4x + 10)}{2} \, dx \\ &= \frac{1}{2} \int 1 \, dx \, – \frac{1}{2} \int \cos(4x + 10) \, dx \\ &= \frac{1}{2}x \, – \frac{1}{2} \cdot \frac{1}{4} \sin(4x + 10) + C, \quad \text{C is a constant} \\ &= \frac{1}{2}x \, -\frac{1}{8} \sin(4x + 10) + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \displaystyle \boxed{\int \sin^{2}(2x + 5) \, dx = \frac{1}{2}x \, – \frac{1}{8} \sin(4x + 10) + C} \)Please let me know in the comments if you find any errors in this solution.