Question

What is `int cos^6x dx`?

Answer 1

`int cos^6 x dx = 5/16 x + 1/4 sin 2x + 3/16 sin 4x -1/48 sin^3 2x`

Explanation:

`int cos^6x dx`

We can rewrite this function as;

`int (cos^2x)^3dx`

Now we can rewrite `cos^2 x` using the half angle formula.

`int(1/2(1+cos(2x)))^3 dx`

If we expand the expression we can get rid of the exponent.

`1/8 int (1 + 3cos(2x) + 3cos^2 (2x) + cos^3 (2x) )dx`

We can split up this expression using the sum rule.

`1/8 int dx + 3/8 int cos (2x) dx + 3/8 int cos^2 (2x) dx + 1/8 int cos^3 (2x) dx`

The first integral is pretty straight forward. For the second, use the substitution `u=2x, du = 2dx`.

`1/8x + 3/16 sin (2x) + 3/8 int cos^2 (2x) dx + 1/8 int cos^3 (2x) dx`

The remaining integrals are a little more involved.

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We can use the half angle formula again to simplify the 3rd integral.

`int cos^2 (2x) dx = int 1/2(1+ cos(4x)) dx`

`= 1/2 int dx + 1/2 int cos(4x) dx`

Use substitution again to solve the second integral.

`1/2 x + 1/8 sin (4x)`

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The last integral should be split up first.

`int cos^3 (2x) dx = int cos^2 (2x) cos (2x) dx`

Now we can use the Pythagorean theorem to replace `cos^2(2x)`.

`int (1-sin^2 (2x)) cos (2x) dx`

`int cos (2x) dx - int sin^2 (2x) cos (2x) dx`

`1/2 sin (2x) - int sin^2 (2x) cos (2x) dx`

To solve the remaining integral, use the substitution `u=sin (2x), du = 2 cos (2x) dx`.

`1/2 sin (2x) - 1/6 sin^3 (2x)`

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Plug the solved integrals into the original expression to get;

`1/8x + 3/16 sin 2x + 3/8(1/2 x + 1/8 sin 4x ) + 1/8(1/2 sin 2x - 1/6 sin^3 2x)`

Simplify by combining like terms.

`5/16 x + 1/4 sin 2x + 3/16 sin 4x -1/48 sin^3 2x`