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

Answer 1

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

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.

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Answer 2

Evelyn Agard

The integral of ( 16 \sin^2(x) \cos^2(x) , dx ) is ( \frac{8}{3}x - \frac{4}{3}\sin(4x) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.