How do I evaluate the indefinite integral #intsin^2(2t)dt# ?

Answer 1
#=1/2t-1/8sin4t+c#, where #c# is a constant

Justification,

#=intsin^2(2t)dt#
Using trigonometric identity, #cos2t=1-2sin^2t#
#sin^2t=(1-cos2t)/2#, inserting this value of #sin^2t# in integral, we get
#=int(1-cos4t)/2dt#
#=1/2int1dt-1/2intcos4tdt#
#=1/2t-1/2(sin4t)/4+c#, where #c# is a constant
#=1/2t-1/8sin4t+c#, where #c# is a constant
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Answer 2

To evaluate the indefinite integral of sin^2(2t) dt, you can use trigonometric identities to simplify the expression. One common approach is to use the double angle identity for sine, which states that sin^2(x) = (1 - cos(2x))/2. By applying this identity, you can rewrite sin^2(2t) as (1 - cos(4t))/2. Then, integrate the expression term by term.

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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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