What is int cos^6x dx?

1 Answer
Jan 30, 2016

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.


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)


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)


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