What is the antiderivative of #cos^3(x) sin(x) (root(4){1-cos^4(x)})#?

1 Answer
Jim H
Mar 8, 2015

The antiderivative of #cos^3(x) sin(x) root(4){1-cos^4(x)}#
is #(1/5)(1-cos^4x)^(5/4)+C# which could also be written:

#(1/5)(root(4)(1-cos^4x))^5+C#

The method of solving may depend on your notation:

Method 1:
Let #u=1-cos^4x#
so that #du=-4cos^3x(-sinx)dx=4cos^3xsinxdx#.

Therefore, # cos^3xsinxdx=(du)/4#.

Using integral notation for the antiderivative, we can write:

#intcos^3(x) sin(x) (root(4){1-cos^4(x)})dx#

#=int(1-x^4)^(1/4)cos^3(x) sin(x)dx=intu^(1/4)(du)/4#

#=1/4intu^(1/4)du=1/4(u^(5/4))/(5/4)+C=(1/5)(1-cos^4x)^(5/4)+C#

Method 2:
If you do not recognize the above notation, try this:

Let #g(x)=1-x^4#. If we took the derivative of #g(x)# to a power, we would need to multiply by #g'(x)# which is #4cos^3xsinx#.
(That is what the chain rule tells us.)

The function you asked about is almost #(g(x))^(1/4)g'(x)#.
It is exactly #(1/4)(g(x))^(1/4)g'(x)#.
So the antiderivative is #(1/4) (g(x))^(5/4)/(5/4)+C#

which is the same as #(1/5)(1-cos^4x)^(5/4)+C#

As always, you should check the answer by differentiating.

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