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What is #int 16sin^2 xcos^2 x dx #?

Answer 1

Christopher Allers

# int \ 16sin^2x cos^2x \ dx = 2x - 1/2sin4x + C #

Our goal is to assess the integral:

# I = int \ 16sin^2x cos^2x \ dx #

Making use of the identity

# sin 2A-= 2sinAcosA #

We are able to write:

# I = int \ 4*4*(sinxcosx)^2 \ dx # # \ \ = int \ 4(2sinxcosx)^2 \ dx # # \ \ = int \ 4(sin2x)^2 \ dx # # \ \ = int \ 4 sin^2 2x \ dx #

We then employ the identity:

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

Thus that we can write:

# I = int \ 4 sin^2 2x \ dx # # \ \ = int \ 4 (1/2(1-cos4x)) \ dx # # \ \ = int \ 2-2cos4x \ dx #

which we can easily incorporate:

# I = 2x - (2sin4x)/4 + C # # \ \ = 2x - 1/2sin4x + C #

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Answer 2

Elijah Abram

#int 16sin^2 xcos^2 x dx #

#=2int 2*(2sin xcos x )^2dx #

#=2int 2*sin ^2 2xdx #

#=2int (1-cos 4x)dx #

#=2int dx-2intcos 4xdx #

#=2x-2*(sin4x)/4+c #, where c is integration constant

#=2x-1/2(sin4x)+c #

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Answer 3

Scarlett Allison

The integral of (16\sin^2(x)\cos^2(x) , dx) is (\frac{8}{3}x - \frac{1}{3}\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.