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What is the integral of #cos^2(x)#?

Answer 1

#int cos(x)^2 dx = x/2 + 1/4 Sin(2 x)+C#

#cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)# Making #alpha=beta->cos(2alpha) = cos(alpha)^2-sin(alpha)^2# but #cos(alpha)^2+sin(alpha)^2=1# then #cos(alpha)^2=( 1+cos(2 alpha))/2# so #int cos(x)^2dx = int( 1+cos(2 x))/2dx = 1/2int dx + 1/2intcos(2x)dx# Finally #int cos(x)^2 dx = x/2 + 1/4 Sin(2 x)+C#

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Answer 2

Emma Aldridge

The integral of cos^2(x) dx is (1/2)(x + sin(2x)) + 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.