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How do you find the integral of #sin^2(3x)dx#?

Answer 1

Emilia Aguilar

#=1/2x - 1/12sin6x + C#

#int \ sin^2 (3x) \ dx#

small book keeping gesture is to make the sub #u = 3x, du = 3 dx#

#1/3int \ sin^2 (u) \ du#

then we use the cosine double angle formulae

#cos 2A = 1 - 2 sin^2 A#

so # sin^2 A = (1 - cos 2A)/2#

#=1/6int \ 1 - cos 2u \ du#

#=1/6( u - 1/2sin 2u ) + C#

#=1/6( 3x - 1/2sin( 2*3x) ) + C#

#=1/2x - 1/12sin6x + C#

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Answer 2

Axel Alvarez

To find the integral of sin^2(3x)dx, you can use the double angle identity for sine, which states that sin^2(x) = (1 - cos(2x))/2. Substituting 3x for x, we get sin^2(3x) = (1 - cos(6x))/2. Then integrate this expression with respect to x. The integral of (1 - cos(6x))/2 dx equals (1/2)x - (1/12)sin(6x) + C, where C is the constant of integration. Therefore, the integral of sin^2(3x)dx is (1/2)x - (1/12)sin(6x) + C.

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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.