Find the indefinite integral

The Correct Answer and Explanation is:

The correct answer is:

a. (sin⁡(2x)/30)⋅(4−5sin⁡2(2x)+3sin⁡4(2x))+C(\sin(2x)/30) \cdot (4 – 5\sin^2(2x) + 3\sin^4(2x)) + C

Explanation:

To solve the integral

∫cos⁡3(2x)⋅sin⁡2(2x) dx,\int \cos^3(2x) \cdot \sin^2(2x) \, dx,

we start by expressing the odd power of cosine separately as

cos⁡2(2x)⋅cos⁡(2x).\cos^2(2x) \cdot \cos(2x).

Using the Pythagorean identity, we replace cos⁡2(2x)\cos^2(2x) with 1−sin⁡2(2x)1 – \sin^2(2x), and then apply the substitution:

u=sin⁡(2x),du=2cos⁡(2x)dx⇒12du=cos⁡(2x)dx.u = \sin(2x), \quad du = 2\cos(2x) dx \Rightarrow \frac{1}{2}du = \cos(2x) dx.

Substitute into the integral:

∫(1−u2)u2⋅12du=12∫(u2−u4)du.\int \left(1 – u^2\right) u^2 \cdot \frac{1}{2} du = \frac{1}{2} \int (u^2 – u^4) du.

Now integrate term by term:

12(u33−u55)+C.\frac{1}{2} \left( \frac{u^3}{3} – \frac{u^5}{5} \right) + C.

To combine under a common denominator:

12(5u3−3u515)=130u(5u2−3u4)+C.\frac{1}{2} \left( \frac{5u^3 – 3u^5}{15} \right) = \frac{1}{30} u (5u^2 – 3u^4) + C.

Now substitute back u=sin⁡(2x)u = \sin(2x):

sin⁡(2x)30(5sin⁡2(2x)−3sin⁡4(2x))+C.\frac{\sin(2x)}{30} \left(5\sin^2(2x) – 3\sin^4(2x)\right) + C.

Factor out as:

sin⁡(2x)30(4−5sin⁡2(2x)+3sin⁡4(2x))+C.\frac{\sin(2x)}{30} \left(4 – 5\sin^2(2x) + 3\sin^4(2x)\right) + C.

This matches choice a, so it is the correct answer. This technique uses trigonometric substitution to simplify the integrand, a standard approach when handling powers of sine and cosine.