Question

What is `int 16sin^2 xcos^2 x dx`?

Answer 1

`2x - sin(4x)/2 + k` with `k in RR`.

Explanation:

We have to remember a few formulas. Here, we will need `2sin(theta)cos(theta) = sin(2theta)`. We can make it appear easily because we're dealing with the squares of `sin(x)` and `cos(x)` and we're multiplying them by an even number.

`16sin^2(x)cos^2(x) = 4(4cos^2(x)sin^2(x)) = 4(2sin(x)cos(x))^2 = 4(sin(2x))^2`.

So `int16sin^2(x)cos^2(x)dx = 4intsin^2(2x)dx`.

And we know that `sin^2(theta) = (1-cos(2theta))/2` because `cos(2theta) = 1-2sin^2(theta)`, so `sin^2(2x) = (1 - cos(4x))/2`.

Hence the final result : `4intsin^2(2x) = 4int(1 - cos(4x))/2dx = 4intdx/2 - 4intcos(4x)/2dx = 2x - 2intcos(4x)dx = 2x + c - 2sin(4x)/4 + a` with `a,c in RR`. Let's say `k = a + c`, hence the final answer.