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How do you find the antiderivative of #int (cos^2x-sin^2x)dx# from #[0,pi/6]#?

Answer 1

Aurora Alvarado

#int_0^(pi/6)(cos^2x-sin^2x)dx=sqrt3/4#

#int_0^(pi/6)(cos^2x-sin^2x)dx#

#=int_0^(pi/6)(cos2x)dx#

#=[1/2sin2x]_0^(pi/6)#

#=1/2sin2(pi/6)-1/2sin2(0)#

#=sqrt3/4-0=sqrt3/4#

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Answer 2

Axel Alvarez

To find the antiderivative of ( \int (\cos^2(x) - \sin^2(x)) , dx ) from ( 0 ) to ( \frac{\pi}{6} ), you can use the trigonometric identity ( \cos^2(x) - \sin^2(x) = \cos(2x) ). Then integrate ( \cos(2x) ) with respect to ( x ) and evaluate the integral from ( 0 ) to ( \frac{\pi}{6} ). The result is ( \frac{1}{2} \sin(2x) ) evaluated from ( 0 ) to ( \frac{\pi}{6} ), which equals ( \frac{1}{2} \sin(\frac{\pi}{3}) - \frac{1}{2} \sin(0) = \frac{\sqrt{3}}{4} ).

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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.