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

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Answer 1 · 2018-04-10T12:46:48

#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 #

Answer 2 · 2018-04-10T12:47:00

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

We want to evaluate the integral:

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

Using the identity:

# sin 2A-= 2sinAcosA #

We can write:

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

Next we use the identity:

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

So we can write:

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

Which we can readily integrate:

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