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What is the integral of #int sin^2 (x).cos^2 (x) dx #?

Answer 1

Emily Ammerman

#intsin^2xcos^2xdx=x/8-(sin4x)/32+c#

As #sin2x=2sinxcosx#

#intsin^2xcos^2xdx=1/4int(4sin^2xcos^2x)dx#

= #1/4intsin^2(2x)dx#

= #1/4int(1-cos4x)/2dx#

= #x/8-1/8intcos4xdx#

= #x/8-1/8xx(sin4x)/4+c#

= #x/8-(sin4x)/32+c#

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Answer 2

Ezra Adams

The integral of ( \sin^2(x) \cdot \cos^2(x) ) with respect to ( x ) is:

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