How do you integrate #int sin^(4) (2x) #?

Answer 1

#intsin^4(2x)dx=3/8x-1/8sin(4x)+1/64sin(8x)+C#

#I=intsin^4(2x)color(red)(dx)#

When dealing with even powers of sine and cosine, the cosine double-angle identity is a great tool to use if substitution is not an option.

The form of the cosine double-angle identity involving sine is #cos(2theta)=1-2sin^2(theta)#. This can be solved for #sin^2(theta)# as #sin^2(theta)=(1-cos(2theta))/2#.

Keep in mind that the following forms are comparable:

#sin^2(theta)=(1-cos(2theta))/2" "=>" "color(blue)(sin^2(2x)=(1-cos(4x))/2#

The cosine function's argument must contain twice as much as the sine function's. Next, we can rewrite the integrand using the blue identity:

#I=int(sin^2(2x))^2dx=int((1-cos(4x))/2)^2dx#
From here, expand #(1-cos(4x))^2#:
#I=1/4int(1-2cos(4x)+cos^2(4x))dx#
#I=1/4intdx-1/2intcos(4x)dx+1/4intcos^2(4x)dx#
The first two integrals are solved fairly easily. The second one can be solved using the chain rule in reverse (with the substitution #u=4x=>du=4color(white).dx#).
#I=1/4x-1/8int4cos(4x)dx+1/4intcos^2(4x)dx#
#I=1/4x-1/8sin(4x)+1/4intcos^2(4x)dx#
To resolve this even power of cosine, use another form of the cosine double-angle identity: #cos(2theta)=2cos^2(theta)-1#. This shows that #cos^2(theta)=(1+cos(2theta))/2#.

Which is comparable to:

#cos^2(theta)=(1+cos(2theta))/2" "=>" "color(green)(cos^2(4x)=(1+cos(8x))/2)#

Changing this to:

#I=1/4x-1/8sin(4x)+1/4int(1+cos(8x))/2dx#

which can subsequently be divided and combined as before:

#I=1/4x-1/8sin(4x)+1/8intdx+1/8intcos(8x)dx#
#I=1/4x-1/8sin(4x)+1/8x+1/64int8cos(8x)dx#
#I=3/8x-1/8sin(4x)+1/64sin(8x)+C#
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Answer 2

To integrate ( \int \sin^4(2x) ), you can use the power-reducing identity for sine, which states that ( \sin^2(x) = \frac{1 - \cos(2x)}{2} ).

So, we can rewrite ( \sin^4(2x) ) as ( (\sin^2(2x))^2 ).

Substituting ( \sin^2(2x) = \frac{1 - \cos(4x)}{2} ), we get:

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

Expand and simplify the expression, then integrate term by term.

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

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

To integrate ( \frac{1}{4} - \frac{1}{2}\cos(4x) + \frac{1}{4}\cos^2(4x) ), you integrate each term separately:

[ \int \frac{1}{4} , dx - \int \frac{1}{2}\cos(4x) , dx + \int \frac{1}{4}\cos^2(4x) , dx ]

[ = \frac{1}{4}x - \frac{1}{8}\sin(4x) + \frac{1}{4} \int (1 + \cos(8x)) , dx ]

[ = \frac{1}{4}x - \frac{1}{8}\sin(4x) + \frac{1}{4}\left(x + \frac{1}{8}\sin(8x)\right) + C ]

[ = \frac{1}{4}x + \frac{1}{4}x - \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C ]

[ = \frac{1}{2}x - \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C ]

So, ( \int \sin^4(2x) , dx = \frac{1}{2}x - \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + 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.

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.

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.

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.

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