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How do you find the antiderivative of #sin^4(x)#?

Answer 1

Isaiah Allen

You could apply double angle power reduction formulas.

Because #cos 2 theta = 1-2sin^2 theta = 2cos theta -1#, we get

#sin^2theta = 1/2 (1-cos2 theta)# and #cos^2theta = 1/2 (1+cos2 theta)#

#int sin^4x dx = int (sin^2x)^2 dx#

#color(white)"sssssssssss"# # = int [1/2(1-cos2x)]^2 dx#

#color(white)"sssssssssss"# # = 1/4 int (1-2cos2x+ cos^2 2x) dx#

#color(white)"sssssssssss"# # = 1/4 int (1-2cos2x+ 1/2(1+cos4x)) dx#

#color(white)"sssssssssss"# # = 1/4 int (3/2-2cos2x+ 1/2 cos4x) dx#

#color(white)"sssssssssss"# # = 1/4 (3/2 x - sin2x+ 1/8 sin4x) +C#

#color(white)"sssssssssss"# # = 3/8 x- 1/4 sin2x+ 1/32 sin4x +C#

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Answer 2

Charles Arocha

To find the antiderivative of ( \sin^4(x) ), you can use the power-reducing formula for sine raised to an even power:

[ \sin^2(x) = \frac{1}{2} - \frac{1}{2} \cos(2x) ]

Using this formula twice, you get:

[ \sin^4(x) = \left(\frac{1}{2} - \frac{1}{2} \cos(2x)\right)^2 ]

Expand and simplify this expression. Then integrate term by term. You'll eventually get the antiderivative of ( \sin^4(x) ).

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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.