How do you find the integral of #(sin^2(6x))(cos^2(6x))dx#?

Answer 1

You can rewrite this as follows

#int(sin^2(6x))(cos^2(6x))dx=int (1/22sin(6x)*cos(6x))^2dx= int 1/4(sin12x)^2 dx#

Remember that

#cos2x=cos^2x-sin^2x=1-2sin^2x=> sin^2x=1/2*(1-cos2x)#

Hence #1/4int (sin^2 12x)dx=int 1/8(1-cos24x)dx=x/8-1/192*sin(24 x)+c#
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Answer 2

To find the integral of ((\sin^2(6x))(\cos^2(6x))dx), you can use trigonometric identities. Specifically, you can use the identity (\sin^2(x) = \frac{1 - \cos(2x)}{2}) and (\cos^2(x) = \frac{1 + \cos(2x)}{2}) to express (\sin^2(6x)) and (\cos^2(6x)) in terms of (\cos(12x)). After substitution, the integral becomes (\int \frac{1 - \cos(2 \cdot 6x)}{2} \cdot \frac{1 + \cos(2 \cdot 6x)}{2} , dx). Then, expand the expression and integrate each term individually. The final integral will be (\frac{1}{4}x - \frac{1}{48}\sin(12x) + 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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