Question #8bc6b

2 Answers
Jan 6, 2018

#intcos^3xdx=sinx-sin^3x/3+"C"#

Explanation:

Rewrite the integral as #intcos^2xcosxdx#

Use the identity #sin^2x+cos^2x=1# and rewrite as:

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

Are integral is now

#int(1-sin^2x)(cosx)dx#

Apply u-substitution

Let #u=sinx->du=cosxdx#

#int(1-u^2)du#

#=u-u^3/3#

Substitute back for #u#

#=sinx-sin^3x/3+"C"#

Jan 6, 2018

#color(red)(sinx - 1/3 sin^3 x + c) or color(blue)(1/12 sin3x + 3/4 cosx + c#

Both are equal!

Explanation:

Method 1 - #u# - substitution

Lets make a #u# substitution...

let # u = sinx #

then #du = color(red)(cosx dx #

Lets re-write this integral: #int cos^3 x dx -= int cos^2 x * color(red)(cosx dx#

Now we can transform this integral...

# => int cos^2 x * du #

#=> u^2 = sin^2 x #

#=> 1 - u^2 = cos^2 x # #" Using our trig identities..." #

#=> int 1-u^2 du = u - 1/3 u^3 + c " Using reverse power rule..." #

Substituting #u= sinx # back:

#color(blue)(int cos^3 x dx = sinx - 1/3 sin^3 x + c_0 #

#c_0 - "Constant" #

Method 2 - complex numbers

This method invloves the use of complex numbers

We must utilise: #color(orange)( omega + 1 / omega = 2cosx #

Where #omega = cosx + i sinx #

Hence #(omega + 1 / omega) ^3 = 2^3 cos^3 x #

#=> 8cos^3 x = omega^3 + (3omega^2 * 1/omega) + (3omega * 1/omega^2 ) + 1/omega^3 #

#=> 8cos^3 x = omega^3 + 1/omega^3 + 3( omega + 1/omega) #

We know #2cosn x = omega^n + 1/omega^n #

#=> 8cos^3x = 2cos3x + 6cosx #

#=> cos^3x = 1/4(cos3x + 3 cosx) #

Hence our integral becomes: #1/4 int cos3x + 3sinx dx #

#= color(blue)(1/12 sin3x + 3/4 cosx + c_1 #

#c_1 - "Constant" #

Both of these solutions are equal...

#color(red)(sinx - 1/3 sin^3 x + c) or color(blue)(1/12 sin3x + 3/4 cosx + c#:

desmos.com