Question

Find the integrals of the function: $sin^2 (2x + 5)$

Answer 1

The correct answer is: $\frac{1}{2}x-\frac{1}{8}sin(4x+10)+C$
$sin^2 (2x + 5) = \frac{1-cos\,2(2x + 5)}{2} = \frac{1-cos(4x + 10)}{2}$
$⇒ \int sin^2(2x + 5)dx = \int \frac{1-cos(4x+10)}{2} dx$
$=\frac{1}{2} ∫1 dx-\frac{1}{2} ∫cos(4x + 10)dx$
$= \frac{1}{2}x-\frac{1}{2}(\frac{sin(4x + 10)}{4})+C$
$=\frac{1}{2}x-\frac{1}{8}sin(4x+10)+C$