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What is the antiderivative of #(sin(x))^2#?

Answer 1

Aurora Alvarado

# = 1/2 (x -1/2 sin 2x) + C#

use the handy double-angle formula

#cos 2A = cos^2 A - sin^2 A# #= 2 cos^2 A -1# #= 1 - 2 sin^2 A#

last one gives us #sin^2 A = 1/2(1- cos 2A)#

#int (sin(x))^2 dx#

# = 1/2 int 1 - cos 2x dx#

# = 1/2 (x -1/2 sin 2x) + C#

which you can flip using the other double-angle formula: #sin 2A = 2 sin A cos A#

# = 1/2 (x - sin x cos x) + C#

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Answer 2

Scarlett Allison

The antiderivative of ( (\sin(x))^2 ) is ( \frac{1}{2}x - \frac{1}{4}\sin(2x) + 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.